Quantizing a two-dimensional Fermion Oscillator

[jostpuur]

Problem:

How do you quantize a two dimensional system defined by the Lagrange's function

<br /> L=\dot{x}y - x\dot{y} - x^2 - y^2 ?<br />

This is a non-trivial task, because the system has some pathology. Classically the equations of motion are

<br /> \dot{x}(t) = y(t)<br />
<br /> \dot{y}(t) = -x(t),<br />

and for arbitrary initial configuration x(0), y(0), the solution is

<br /> \left[\begin{array}{c}<br /> x(t) \\ y(t) \\<br /> \end{array}\right]<br /> = \left[\begin{array}{rr}<br /> \cos t &amp; \sin t \\<br /> -\sin t &amp; \cos t \\<br /> \end{array}\right]<br /> \left[\begin{array}{c}<br /> x(0) \\ y(0) \\<br /> \end{array}\right]<br />

Alternatively, the L, EOM, and solution can be written more compactly with complex numbers:

<br /> L=\textrm{Im}(\dot{z}^* z) - |z|^2<br />
<br /> \dot{z} = -iz<br />
<br /> z(t)=e^{-it}z(0)<br />

Usually the equations of motion define the second time derivatives of the coordinates, so that both coordinates and velocities are needed for unique solution. With this system velocities are not independent of the coordinates, and as consequence the usual quantization procedure doesn't really work.

An attempt with the Schrödinger's equation and Hamiltonian:

The Hamiltonian can be solved to be

<br /> H=\frac{\partial L}{\partial\dot{x}}\dot{x} + \frac{\partial L}{\partial\dot{y}}\dot{y} - L = x^2 + y^2 = |z|^2<br />

This looks pretty strange Hamiltonian, but it is in fact one of the most obviously conserving quantities in the system, so it could be considered some kind of energy. The SE is then

<br /> i\partial_t\Psi(t,z) = |z|^2\Psi(t,z),<br />

and solutions are

<br /> \Psi(t,z) = e^{-i|z|^2 t}f(z).<br />

Clearly something is wrong with this, because the quantum mechanical solutions do not give the classical behavior on the classical limit. The problem is that the canonical momenta are

<br /> p_x = \frac{\partial L}{\partial\dot{x}} = y<br /> \quad\quad\quad<br /> p_y = \frac{\partial L}{\partial\dot{y}} = -x<br />

so when we should leave parameters x and y untouched, and substitute p_x\to -i\partial_x and p_y\to -i\partial_y, the coordinates and momenta get confused. I don't really know where those derivative operators should be put.

An attempt with path integrals and action:

Suppose the system goes from z' to z, in time t_1-t_0. We can parametrize a path

<br /> z(t) = \frac{(t_1-t)z&#039; + (t-t_0)z}{t_1-t_0},<br />

and compute the action

<br /> S = \int\limits_{t_0}^{t_1} L(z(t))dt = xy&#039; - yx&#039; - \frac{1}{3}(t_1-t_0)\big( x^2 + xx&#039; + (x&#039;)^2 + y^2 + yy&#039; + (y&#039;)^2\big),<br />

but this cannot be used to define time evolution with

<br /> \Psi(t+\Delta t, z) = N \int dx&#039;\;dy&#039;\; e^{iS}\Psi(t,z&#039;)\;+\; O(\Delta t^2)<br />

as usual, because there is no

<br /> \propto \quad\frac{1}{t_1-t_0},\quad\quad\textrm{or}\quad \frac{1}{(t_1-t_0)^2}<br />

kind of terms in the S as usual. Such terms are necessary to make actions for large spatial transition in small time to become infinite, and to produce necessary oscillation in the path integral.

At this point I'm out of ideas.

Motivation

This is not an unphysical example. Even though physical systems usually have EOM containing second order time derivatives, this is not always the case, and the classical Dirac field is the most obvious counter example, since Dirac equation contains only first order derivatives. Also, any initial \psi(0,x) alone always fixes the time evolution uniquely.

One question that stroke me already in the beginning of the studies of QFT was, that if the quantization of the Klein-Gordon field is based on the quantization of harmonic oscillators, then what is the quantization of the Dirac's field based on? This question is easily answered by writing the Lagrangian

<br /> L = \int d^3x\; \textrm{Re}\Big(i\overline{\psi}(x)\gamma^{\mu}\partial_{\mu}\psi(x) - m\overline{\psi}(x)\psi(x)\Big)<br />

in terms of the Fourier coefficients of psi. The answer is

<br /> \int\frac{d^3p}{(2\pi)^3}\Big(\textrm{Re}(i\psi_p^{\dagger}\partial_0 \psi_p) - \boldsymbol{p}\cdot(\overline{\psi}_p\boldsymbol{\gamma}\psi_p) - m\overline{\psi}_p \psi_p\Big)<br />

So the fermion analogy to the harmonic oscillator is

<br /> L = \textrm{Im}(\dot{z}^{\dagger} z) - \boldsymbol{a}\cdot(\overline{z}\boldsymbol{\gamma} z) - b\overline{z} z,\quad\quad\quad z(t)\in\mathbb{C}^4<br />

The classical behavior of this system is straightforward to solve. It is more complicated than the simple example in the beginning of the post, but it has the same basic properties, and in particular the EOM contains only first order time derivatives.

So ultimately, I would like to understand how to quantize this fermion oscillator, but right now I'm more interested in the simpler example, because the main difficulty is already present there.

Already known:

There is no need to preach me that it is sufficient to take the harmonic oscillator, and replace the commutation relations of the operators by anti-commutation relations. I am fully aware that this is how Dirac field is usually made to work. But besides the abstract properties of the raising and lowering operators a^{\dagger} and a of the harmonic oscillator, these operators also have very explicit expressions in terms of the operators x and -i\partial_x. I would like to have something similarly explicit for the fermion oscillator too.

 

[jostpuur]

Even though I don't understand this thing fully, it is easy to get some kind of vague intuitive idea, how systems where classical behavior is given by second order time differential equations, are described quantum mechanically by formulation where some operators satisfy some commutation relations, and on the other hand systems where classical behavior is given by first order time differential equations, are described quantum mechanically by formulation where some operator satisfy anti-commutation relations.

I can almost feel, how there is something elegant only waiting to be understood! But I have no knowledge what it could be...

Worse yet, I don't know if somewhere there already is a book that would explain this all, or if physicists simply are not even interested in this thing, because doing stuff with postulated operators works well enough.

 

[reilly]

If Jz is the z component of angular momentum, then L = Jz - r^^2, and H=Jz +r^^2, where r^^2 =X^^2 +y^^2. Jz and r^^2 commute. Jz is cool, but what about r^^2?.Convert to momentum space, so that r^^2 --> -( d/dPx (d/dPx + d/dPy(d/Pdy), or minus the two dimensional Laplacian in momentum space. So, you have a standard problem in p space, which is a tricky one in x space. In fact, this problem is pretty much that of the Zeeman effect in momentum space.( All masses are set to 1.)
Regards,
Reilly Atkinson

 

[olgranpappy]

...
<br /> H=\frac{\partial L}{\partial\dot{x}}\dot{x} + \frac{\partial L}{\partial\dot{y}}\dot{y} - L = x^2 + y^2 = |z|^2<br />

This looks pretty strange Hamiltonian, but it is in fact one of the most obviously conserving quantities in the system, so it could be considered some kind of energy...

and it's non-negative, which is nice... but are there any other conserved quantities? the total momentum isnt...

 

[Rainbow Child]

The Lagrangian you posted is non-regular. In order for someone to pass from the Lagrangian description to the Hamiltonian one it must hold

det(\frac{\partial^2\, \mathcal{L}}{\partial\,\dot{q}^i\,\dot{q}^j}) \neq 0 \quad (1)

It is easy to see that at your example, you can not produce the right equations of motion with the Hamiltonian you have. If equation (1) is violated the system is called non-regular because you can not solve the equations

p_i=\frac{\partial \,\mathcal{L}}{\partial\dot{q}^i}

for the velocities, so the imply conditions on some of the (p^i,q^j), which are called primary constraints.

In such cases there is a algrorithm which enables you to write down the Hamiltonian, called extended Hamiltonian, the Dirac-Bergmann algorithm.
With this method you define the Dirac brackets, an extension of Poisson brackets in order to produce the correct equations of motion.

The most striking result is for your example that the Dirac brackets for the phace variables (x,y) does not commute, i.e.

\{x,y\}_{Dirac}=-2 \quad (!)

I can almost feel, how there is something elegant only waiting to be understood! But I have no knowledge what it could be...

Worse yet, I don't know if somewhere there already is a book that would explain this all, or if physicists simply are not even interested in this thing, because doing stuff with postulated operators works well enough.

Of course the physicists are interested! The most known example of this behavior is General Relativity. Actual, this was the starting point for Dirac.

The best book is Dirac's original lectures, i.e.

P. A. M. Dirac, "Lectures on Quantum Mechanics" Belfer Graduate School of Science, Yeshiva University, New York (1964);
P. A. M. Dirac, Proc. Roy. Soc. A246, 326 (1958a)
P. A. M. Dirac, Canad. J. Math. 2, 129 (1950)

For a more extensive treatment (e.g. Hamiltonian/Lagrangian Formalism, Path-Integrals) with a great variety of applications (Gravitation, Yang-Mills Theories, Strings, e.t.c.), you could try

K. Sundermeyer, "Constrained Dynamics", Spinger-Verlag (1982)

and for a more advanced textbook (difficulty\rightarrow \infty)

Henneaux, Marc and Teitelboim, Claudio, Quantization of Gauge Systems. Princeton University Press, (1992)

 

[reilly]

There is nothing peculiar about the problem discussed here.; First, go revisit Goldstein's text.

That is 1. look up angular momentum.

2.then consider polar coordinates, and uniform circular motion -- just to get the basic physics. And, think about the utility of polar coordinates for this problem.

3. Next, look in Goldstein -- or Lanczos -- re canonical transformations that interchange q and p. Note that the oscillator Hamiltonian is invariant under
p <--> q -- H must, of course be properly parameterized.(In QM this is equivalent to going from the configuration space representation to the momentum representation.)This is about as sophisticated as you need to get, and, probably more than you need to.)

4. Review my post above for the nitty-gritty of the problem.

I wish I'd known this problem when I was teaching mechanics. However I'd go to 3 dimensions, and make the angular momentum term something like a "complete" Zeeman-like interaction,

H == B dot L + R dot R;

B is the magnetic field, L is the orbital angular momentum, and R is the 3-D position vector of the particle..

That is, this problem would be great for a final exam, for albeit a very sophisticated undergraduate course, or pretty straightforward for a graduate mechanics course.

This is a very good problem. Also, would be terrific as a question for a PhD oral exam -- either as a classical or a quantum. problem, or both. In this question
I'd definitely ask about the phase relationships -- in order to test understanding of momenta, which, initially seem to be a bit peculiar.

Keep it simple -- this will be particularly important for a QM interpretation. You don't need fancy for this one.

Regards, Reilly Atkinson

 

[jostpuur]

If Jz is the z component of angular momentum, then L = Jz - r^^2, and H=Jz +r^^2, where r^^2 =X^^2 +y^^2. Jz and r^^2 commute. Jz is cool, but what about r^^2?.Convert to momentum space, so that r^^2 --> -( d/dPx (d/dPx + d/dPy(d/Pdy), or minus the two dimensional Laplacian in momentum space. So, you have a standard problem in p space, which is a tricky one in x space. In fact, this problem is pretty much that of the Zeeman effect in momentum space.( All masses are set to 1.)
Regards,
Reilly Atkinson

I understood nothing out of this! Could it be, that you just took a quick glance on my post and assumed it was something else that you already knew well?

and it's non-negative, which is nice...

Now when you mentioned the non-negativeness, I must note that it doesn't necessarily mean what it might seem to mean. If we instead started with a Lagrangian

<br /> L = \dot{x}y - x\dot{y} + x^2 + y^2,<br />

the Hamiltonian would be

<br /> H=-|z|^2,<br />

but only difference in the solutions would be, that the oscillation goes in the different direction:

<br /> z(t) = e^{it}z(0)<br />

This system doesn't work very intuitively.

The Lagrangian you posted is non-regular. In order for someone to pass from the Lagrangian description to the Hamiltonian one it must hold

det(\frac{\partial^2\, \mathcal{L}}{\partial\,\dot{q}^i\,\dot{q}^j}) \neq 0 \quad (1)

It is easy to see that at your example, you can not produce the right equations of motion with the Hamiltonian you have. If equation (1) is violated the system is called non-regular because you can not solve the equations

p_i=\frac{\partial \,\mathcal{L}}{\partial\dot{q}^i}

for the velocities, so the imply conditions on some of the (p^i,q^j), which are called primary constraints.

Good to see that there is some standard knowledge related to this.

In such cases there is a algrorithm which enables you to write down the Hamiltonian, called extended Hamiltonian, the Dirac-Bergmann algorithm.
With this method you define the Dirac brackets, an extension of Poisson brackets in order to produce the correct equations of motion.

The most striking result is for your example that the Dirac brackets for the phace variables (x,y) does not commute, i.e.

\{x,y\}_{Dirac}=-2 \quad (!)

hmhmh... okey, I don't fully understand this yet. But this is classical stuff anyway? Does this help with the quantization problem too?

Of course the physicists are interested!

Unless they've been brainwashed to the "don't think about it anymore, it already works!"-attitude

The most known example of this behavior is General Relativity.

I don't know GR yet. I'm taking the first course on it this spring. But are you saying, that the quantization problem of GR has something to do with the problem of my OP?

 

[reilly]

jostpuur Ok, let's try Resnick and Halliday, Chapters 11, 12, 13 re rotational motion, and angular momentum. Once you get through this, the simplicity of your problem should become evident.(Most any freshman text should do the job.)

I took a serious look at your problem. In fact it threw me off for a moment. But upon noting that the first two terms define the z component of angular momentum for a unit mass particle, and that this term commutes with r^^2, it all fell into place.

Also, the QM solution factors into two independent terms; W*R, where W, is an eigenstate of Lz, and R is an eigenstate of r^^2, best expressed in momentum space.

With all due respect, if my discussion does not make sense, then you need to study a good bit more QM. It's just not a hard problem.

Or, if you can show me the error of my ways, by all means do so.

By the way, your Hamiltonian in your first post is incorrect. With unit mass there is no practical difference between p and v. That your H is incorrect should be evident from the EOM earlier in that same post. The H is simply Lz + r^^2 in a z=const. plane.
Regards,
Reilly Atkinson

Note also that the problem will be much simpler when expressed in polar coords.

 

[olgranpappy]

jostpuur Ok, let's try Resnick and Halliday, Chapters 11, 12, 13 re rotational motion, and angular momentum. Once you get through this, the simplicity of your problem should become evident.(Most any freshman text should do the job.)

I took a serious look at your problem. In fact it threw me off for a moment. But upon noting that the first two terms define the z component of angular momentum for a unit mass particle, and that this term commutes with r^^2, it all fell into place.

Also, the QM solution factors into two independent terms; W*R, where W, is an eigenstate of Lz, and R is an eigenstate of r^^2, best expressed in momentum space.

With all due respect, if my discussion does not make sense, then you need to study a good bit more QM. It's just not a hard problem.

with all due respect. you don't make sense to me either.

 

[jostpuur]

Reilly, your idea does make sense to me now. It was just that I was thinking about my own problems, and you took a different direction right in the beginning. So you were talking about a system described by a Hamiltonian

<br /> H = \boldsymbol{x}\times\boldsymbol{p} + |\boldsymbol{x}|^2?<br />

(With two dimensional vector notation \boldsymbol{x}\times\boldsymbol{p}=x_1 p_2 - x_2 p_1)

By the way, your Hamiltonian in your first post is incorrect.

The calculation is quite short. I don't think there is a mistake

<br /> L = \dot{x}y - x\dot{y} - x^2 - y^2<br />

<br /> H = \frac{\partial L}{\partial\dot{x}}\dot{x}\; +\; \frac{\partial L}{\partial\dot{y}}\dot{y} \;-\; L = y\dot{x} \;+\; (-x)\dot{y} \;-\; \big(\dot{x}y \;-\; x\dot{y}\; -\; x^2 \;-\; y^2\big) = x^2 + y^2<br />

This is simply a different system than the one described by a Hamiltonian

<br /> H = xp_y - yp_x + x^2 + y^2<br />

 

[Rainbow Child]

hmhmh... okey, I don't fully understand this yet. But this is classical stuff anyway? Does this help with the quantization problem too?

Yes it does! The quantization procesure, runs under the postulate, that you use the Dirac's brackets instead of the Poisson's brackets to formulate the commutators.
In your problem we have

\{x,y\}_{Dirac}=-\frac{1}{2},\,\{x,p_x\}_{Dirac}=\{y,p_y\}_{Dirac}=\frac{1}{2}

thus

[\hat{x},\hat{y}]=-\frac{i\,\hbar}{2},\,[\hat{x},\hat{p}_x]=[\hat{y},\hat{p}_y]=\frac{i\,\hbar}{2}

I don't know GR yet. I'm taking the first course on it this spring. But are you saying, that the quantization problem of GR has something to do with the problem of my OP?

I am saying that the Lagrangian or GR in non-regular, so in order to pass from the Lagrangian to the Hamiltonian you have to consider Dirac's algorithm (though in standard textbooks this is not explicity said). And one way to the quantum GR is this one, called canonical quantization.

 

[jostpuur]

I am saying that the Lagrangian or GR in non-regular, so in order to pass from the Lagrangian to the Hamiltonian you have to consider Dirac's algorithm (though in standard textbooks this is not explicity said). And one way to the quantum GR is this one, called canonical quantization.

("Lagrangian of GR is"?)

Is this the quantization, that leads into the famous non-renormalizable divergences?

 

[Rainbow Child]

("Lagrangian of GR is"?)

\mathcal{L}=\int R\,\sqrt{-g}\,d^4\,x

where R is the Ricci scalar and g the determinant of the metric tensor.

Is this the quantization, that leads into the famous non-renormalizable divergences?

Yes!

 

[jostpuur]

Hehe. Sorry for unclarity. I intended to point out your probable typos

the Lagrangian or GR in non-regular

Yes!

Good piece of information to know.

 

[samalkhaiat]

How do you quantize a two dimensional system defined by the Lagrange's function

<br /> L=\dot{x}y - x\dot{y} - x^2 - y^2 ?<br />

This is a non-trivial task, because the system has some pathology. Classically the equations of motion are

<br /> \dot{x}(t) = y(t)<br />
<br /> \dot{y}(t) = -x(t),<br />

This system represents a 2-dimensional harmonic oscillator;

<br /> \ddot{x} + x = 0<br />
<br /> \ddot{y} + y = 0<br />

Classically and quantum mechanically the system can be solved using anyone of the following Lagrangians;
<br /> L_{0} = \frac{1}{2} \left( \dot{x}^{2} + \dot{y}^{2} \right) - \frac{1}{2} \left( x^{2} + y^{2} \right)<br />
<br /> L_{1} = \dot{x} \dot{y} - xy<br />
or
<br /> L_{2} = \frac{1}{2} \left( \dot{x}^{2} - \dot{y}^{2} \right) - \frac{1}{2} \left( x^{2} - y^{2} \right)<br />

This system represents a bosonic oscillator, because the coordinates x and y are (even) commuting variables; xy = yx. In order to describe a "classical Fermi oscillator", your coordinates x & y need to take values in a Grassmann Algebra,i.e., anticommuting numbers; xy = -yx.
I think the following "preaching" is required here:
As you know, the defining property of fermions is the antisymmetry of manny-particle states under the exchange of any 2 particles. In the context of QFT, this property follows from anticommutation relations between the mode operators a(p)and a^{\dagger}(p). However, the canonical quantization of a field taking its values in the set of real or complex numbers can lead only to commutation relations as opposed to anticommutation relations. Stated differently, a given fermionic mode cannot hold more than one particle and consequently a fermion field cannot have a macroscopic value, i.e., its classical limit does not exist in terms of ordinary (or commuting) real or complex numbers! However, a "classical" description of Fermi fields can be given in terms of Grassmann (or anticommuting) numbers. This is because Grassmann numbers correspond to the classical limit \hbar \rightarrow 0 of an anticommutator in the quantum theory;

\{\psi_{i}(x) , \psi^{\dagger}_{j}(y) \} = \hbar \delta_{ij} \delta ( x - y)

We apply to Grassmann numbers the same canonical formalisim as for ordinary real or complex variables, except that their anticommuting property forbids the existence in the Lagrangian of terms quadratic in derivatives.

Let us consider a discrete set of complex Grassmann variables X_{i}(t) and \bar{X}_{i}(t) (i = 1,2,...,n) with the Lagrangian

L = i \bar{X}_{i} M_{ij} \dot{X}_{j} - V( X,\bar{X})

where M is Hermitian.
The Euler-Lagrange equations (in matrix form) are

\dot{X} = - i M^{-1} \frac{\partial V}{\partial \bar{X}}

NOW COMES THE IMPORTANT BIT:

These classical equations are recovered in the quantum case (operator form) from the Heisenberg equation \dot{X} = [iH,X] provided we use the following Hamiltonian and anticommutation relations;

<br /> H = V(X, \bar{X})<br />
<br /> \{ X_{i} , X_{j} \} = \{ X^{\dagger}_{i} , X^{\dagger}_{j} \} = 0<br />
<br /> \{ X_{i} , X^{\dagger}_{j} \} = M^{-1}_{ij}<br />

wherein X_{i} and X^{\dagger}_{i} are now the quantum operators corresponding respectively to the "classical" variables X_{i}, \bar{X}_{i}.

See the classic paper on the subject:

Martin, J.L. "Generalized classical dynamics and the classical analogue of a Fermi oscillator", Proc.Roy.Soc.,1959, A251,536.

See also the very good book:

Henneaux, M. & Teitelboim, C. "Quantization of Gauge Systems" Princeton, 1992.

For your purpose, you only need:
Chapter six ; "Classical Mechanics over a Grassmann Algebra"
and
Chapter seven; "Constrained Systems with Fermi Variables"

regards

sam

 

[samalkhaiat]

If Jz is the z component of angular momentum, then L = Jz - r^^2, and H=Jz +r^^2, where r^^2 =X^^2 +y^^2. Jz and r^^2 commute. Jz is cool, but what about r^^2?.Convert to momentum space, so that r^^2 --> -( d/dPx (d/dPx + d/dPy(d/Pdy), or minus the two dimensional Laplacian in momentum space. So, you have a standard problem in p space, which is a tricky one in x space. In fact, this problem is pretty much that of the Zeeman effect in momentum space.( All masses are set to 1.)

ARE YOU SERIOUS? Before posting your "answer" try to understand the question raised by the OP.
The man asked about the peculiar (classical & quantum mechanical) nature of Fermi fields.

regards

sam

 

[jostpuur]

This system represents a 2-dimensional harmonic oscillator;

<br /> \ddot{x} + x = 0<br />
<br /> \ddot{y} + y = 0<br />

Classically and quantum mechanically the system can be solved using anyone of the following Lagrangians;
<br /> L_{0} = \frac{1}{2} \left( \dot{x}^{2} + \dot{y}^{2} \right) - \frac{1}{2} \left( x^{2} + y^{2} \right)<br />
<br /> L_{1} = \dot{x} \dot{y} - xy<br />
or
<br /> L_{2} = \frac{1}{2} \left( \dot{x}^{2} - \dot{y}^{2} \right) - \frac{1}{2} \left( x^{2} - y^{2} \right)<br />

Wait a minute. We have

<br /> \left\{\begin{array}{l}<br /> \dot{x} = y\\<br /> \dot{y} = -x\\<br /> \end{array}\right.<br /> \quad\implies\quad<br /> \left\{\begin{array}{l}<br /> \ddot{x} = -x\\<br /> \ddot{y} = -y\\<br /> \end{array}\right.<br />

but the converse is not true. So isn't it a bit dangerous to say that the Lagrangian I gave represents a two dimensional harmonic oscillator?

What you explained sounds like that the classical system I wrote down, cannot be quantized directly, but instead we must modify the classical system first, by replacing the real variables x and y by anti-commuting Grassmann numbers, and then we can quantize it.

What Rainbow Child explained, on the other hand, sounded like that we can quantize the system I wrote down without any modifications, and that it would suffice that some Dirac bracket relations would be satisfied.

See the classic paper on the subject:

Martin, J.L. "Generalized classical dynamics and the classical analogue of a Fermi oscillator", Proc.Roy.Soc.,1959, A251,536.

I don't know how to see it. What does the code "Proc.Roy.Soc.,1959, A251,536" mean?

See also the very good book:

Henneaux, M. & Teitelboim, C. "Quantization of Gauge Systems" Princeton, 1992.

https://www.amazon.com/dp/0691037698/?tag=pfamazon01-20

This looks tough stuff. I'm going to be buying some books (Goldstein's and Jackson's books are already decided)... ...once I get some money. I'll be considering this too.

 

[jostpuur]

Another idea: Could it be possible to quantize a system described by

<br /> L=\epsilon(\dot{x}^2 + \dot{y}^2) + \dot{x}y - x\dot{y} - x^2 - y^2<br />

as usual, and then take something reasonable out by setting \epsilon\to 0?

 

[samalkhaiat]

So isn't it a bit dangerous to say that the Lagrangian I gave represents a two dimensional harmonic oscillator?

Your equations of motion represent a 2-D oscillator, your Lagrangian does not represent any system!
It does not represent a Fermi system because it contains the term x^{2} + y^{2}. Also, it cannot represent a bosonic system because it is not quadratic in derivatives.
So, make up your mind! If you want to describe a bosonic system (constrained or not), then you need to include a term like \dot{x}^{2} in your Lagrangian. And to quantize a constrained bosonic system, follow the rule;

\left[ \hat{A} , \hat{B} \right]_{-} = i \hbar \{ A , B \}_{Dirac}

But, if you want to describe a Fermionic system (constrained or not), then your Lagrangian should not contain the term x^{2} + y^{2}, you also need xy + yx = 0. To quantize a constrained Fermi system, the rule to follow is;

\{ \hat{A} , \hat{B} \}_{+} = i \hbar \{ A , B \}_{Dirac}

What you explained sounds like that the classical system I wrote down, cannot be quantized directly, but instead we must modify the classical system first, by replacing the real variables x and y by anti-commuting Grassmann numbers, and then we can quantize it.

I did not talk about constrained systems, I explained the need for Grassmann numbers when constructing a "classical" Lagrangian for Fermi fields. I also explained the need for anticommutators when quantizing such systems. To turn my Lagrangian into a singular one (constrained fermi system), put

<br /> V = 0<br />
<br /> M_{ij} = \delta_{ij}<br />
<br /> \bar{X}_{i} = X_{i}<br />

i.e.,

L = \frac{i}{2}\dot{X}^{i}(t) X^{j}(t) \delta_{ij}

describe a constrained system of real Grassmann variables. The primary (2nd-class) constraints are

P_{i} + \frac{i}{2} \delta_{ij} X^{j} = 0

the basic Dirac brackets are

\{ X_{i} , X_{j} \}_{Dirac} = - i \delta_{ij}

and the quantization rule is

<br /> \{ \hat{X}_{i} , \hat{X}_{j} \}_{+} = i \hbar \{ X_{i} , X_{j} \}_{Dirac} = \hbar \delta_{ij}<br />

What Rainbow Child explained, on the other hand, sounded like that we can quantize the system I wrote down without any modifications, and that it would suffice that some Dirac bracket relations would be satisfied.

He explained correctly how to quantize a constrained bosonic system. However, he applied the correct quantization rules to the wrong Lagrangian of yours.

I don't know how to see it. What does the code "Proc.Roy.Soc.,1959, A251,536" mean?

Go to the library and look for the very famous journal : Proceeding of the Royal Society (London), The year 1959, Journal number A251, Page 536. and you will find that paper! But don't bother because it is as tough as the book I mentioned.

I'm going to be buying some books (Goldstein's and Jackson's books are already decided)...

Yeh, you need to spend time with Goldstein's & Jackson's first!


regards

sam

 

[Rainbow Child]

... Also, it cannot represent a bosonic system because it is not quadratic in derivatives...

Actually we can make his system a physical one!
If we imagine a particle with charge q and mass m moving in the x-y plane along with a strong constant, homogeneous magnetic field towards the z-direction with strength B_o. Let's also assume that there is an external potential V=\frac{m}{2}\,(x^2+y^2).
Then the corresponding Lagragian reads

\mathcal{L}=\frac{m}{2}\,(\dot{x}^2+\dot{y}^2)+\frac{q\,B_o}{2\,c}\,(x\,\dot{y}-y\,\dot{x})-\frac{m}{2}\,(x^2+y^2)

If you demand a very large magnetic field, i.e.

\frac{q\,B_o}{m\,c}\gg 1

we may neglect the kinetic term, arriving to

\mathcal{L}=\frac{q\,B_o}{2\,c}\,(x\,\dot{y}-y\,\dot{x})-\frac{m}{2}\,(x^2+y^2)

which is jostpuur's Lagrangian apart from numerical constants.

 

[samalkhaiat]

Actually we can make his system a physical one!
If we imagine a particle with charge q and mass m moving in the x-y plane along with a strong constant, homogeneous magnetic field towards the z-direction with strength B_o. Let's also assume that there is an external potential V=\frac{m}{2}\,(x^2+y^2).
Then the corresponding Lagragian reads

\mathcal{L}=\frac{m}{2}\,(\dot{x}^2+\dot{y}^2)+\frac{q\,B_o}{2\,c}\,(x\,\dot{y}-y\,\dot{x})-\frac{m}{2}\,(x^2+y^2)

If you demand a very large magnetic field, i.e.

\frac{q\,B_o}{m\,c}\gg 1

we may neglect the kinetic term, arriving to

\mathcal{L}=\frac{q\,B_o}{2\,c}\,(x\,\dot{y}-y\,\dot{x})-\frac{m}{2}\,(x^2+y^2)

Why not neglect the potential term? This way you arrive at more sound dynamical equations

<br /> \ddot{x} = \frac{q B_{0}}{2mc} \dot{y}<br />
<br /> \ddot{y} = - \frac{qB_{0}}{2mc} \dot{x}<br />


regards


sam

 

[Rainbow Child]

Why not neglect the potential term? This way you arrive at more sound dynamical equations

<br /> \ddot{x} = \frac{q B_{0}}{2mc} \dot{y}<br />
<br /> \ddot{y} = - \frac{qB_{0}}{2mc} \dot{x}<br />


regards


sam

Of course you can drop the potential term! But then the system is regular. I was trying to figure out a mechanism in order to give physical meaning to the original Laplacian.

 

[jostpuur]

Your equations of motion represent a 2-D oscillator, your Lagrangian does not represent any system!

Sounds contradictory. The Lagrangian implies these equations of motion.

<br /> L=\dot{x}y-x\dot{y}-x^2-y^2<br />

<br /> \left\{\begin{array}{l}<br /> D_t\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}\\<br /> D_t\frac{\partial L}{\partial \dot{y}} = \frac{\partial L}{\partial y}\\<br /> \end{array}\right.<br /> \quad\implies\quad \left\{\begin{array}{l}<br /> \dot{y} = -\dot{y} - 2x\\<br /> -\dot{x} = \dot{x} - 2y\\<br /> \end{array}\right.<br /> \quad\implies\quad \left\{\begin{array}{l}<br /> \dot{x} = y\\<br /> \dot{y} = -x\\<br /> \end{array}\right.<br />

My problem is, that how do we quantize this system. Classically it seems to be well defined, so the question is reasonable. And as a hint I have, that it probably has something to do with the Fermi stuff.

Yeh, you need to spend time with Goldstein's & Jackson's first!

Hehe... I have of course already studied mechanics and electromagnetism. It's just that probably not from the best possible sources. I want to have these famous books, and see what I've missed. The Weinberg's vol1 was another one, which I've pretty much decided.

 

[samalkhaiat]

I was trying to figure out a mechanism in order to give physical meaning to the original Laplacian.

But you lost the dynamical system in the process. You see, when we write

L_{1} = \dot{r}^{2} + \frac{q}{mc} \vec{B} . ( r \times \dot{r} ) - r^{2}

or

L_{2} = \dot{r}^{2} + \frac{q}{mc} \vec{B} . ( r \times \dot{r} )

we mean to say that, even in the absence of the interaction, L_{1} & L_{2} describe well defined dynamical systems,i.e., 2nd order differential equations (dynamics) can still be obtained from L_{1} and L_{2} with the field B switched off. This fact is not satisfied by your version of the original Lagrangian;

L_{3} = \frac{q}{mc} \vec{B} . ( r \times \dot{r} ) - r^{2}

Clearly, No dynamical system can be obtained from L_{3} when B = 0

So no physical meaning can be attached to the original Lagrangian. Post #19 was all about pointing out this fact. The dynamics of any Bosonic system ( constrained or regular) is determined by Lagrangians that are quadratic in velocities.

regards

sam

 

[samalkhaiat]

My problem is, that how do we quantize this system. Classically it seems to be well defined, so the question is reasonable.

Realy? And what are "your" criteria for a "classically well defined dynamical system"?

And as a hint I have, that it probably has something to do with the Fermi stuff.

Clearly you did not understand what I said in post #19! Look, if your Lagrangian was

L = x \dot{y} - y \dot{x}

with

xy + yx = 0

then yes, it ,definitely, describes constrained classical "Fermi stuff".

You created this thread and gave it the title "fermion oscillator", yet you don't seem to know the difference between Fermi and Bose dynamics.
You wrote an incorrect Bosonic Lagrangian and asked us to help you quantize that wrong Lagrangian! You also asked us to obtain information from the wrong Bosonic Lagrangian and use that information to explain Fermion oscillator! These requests of yours are certainly meaningless!

So, to understand the issues of your thread, you need to gain some knowledge about

1) the even (commuting) classical variables of Bose.
2) the odd (anticommuting) classical variables of Fermi.
3) Bosonic Lagrangians (regular or singular) are quadratic in derivatives.
4) Fermionic Lagrangians (regular or singular) are 1st order in derivatives.
5) special treatement is needed when dealing with singular Lagrangians which describe constrained (Fermi or Bose) systems

regards

sam

 

[jostpuur]

So if I ask a question "How do you quantize the system L=\dot{x}y-x\dot{y}-x^2-y^2?", your answer is "No, that is wrong. You have to quantize something else."

 

[samalkhaiat]

So if I ask a question "How do you quantize the system L=\dot{x}y-x\dot{y}-x^2-y^2?", your answer is "No, that is wrong. You have to quantize something else."

Your so-called "SYSTEM" defines no classical dynamics. Therefore, it is MEANINGLESS to try to quantize it.
Howmany times do I need to tell you this: What you called "Lagrangian" is not a Lagrangian because it represents no dynamical system. So, my answer allways was and still is ; write a correct Lagrangian first!


sam

 

[jostpuur]

Your so-called "SYSTEM" defines no classical dynamics. Therefore, it is MEANINGLESS to try to quantize it.
Howmany times do I need to tell you this: What you called "Lagrangian" is not a Lagrangian because it represents no dynamical system. So, my answer allways was and still is ; write a correct Lagrangian first!

The Lagrangian

<br /> L=\dot{x}y - x\dot{y} - x^2 - y^2 ?<br />

with the action principle, defines a system whose equations of motion are

<br /> \dot{x}(t) = y(t)<br />
<br /> \dot{y}(t) = -x(t),<br />

and whose solutions are

<br /> \left[\begin{array}{c}<br /> x(t) \\ y(t) \\<br /> \end{array}\right]<br /> = \left[\begin{array}{rr}<br /> \cos t &amp; \sin t \\<br /> -\sin t &amp; \cos t \\<br /> \end{array}\right]<br /> \left[\begin{array}{c}<br /> x(0) \\ y(0) \\<br /> \end{array}\right]<br />

I don't understand why this would not be a well defined classical system.

 

[samalkhaiat]

I hope this is not a waste of time!

The calculus of variations can be applied to any function, for example

L = t \dot{x} + \sin{x}

leads to the E-L equation \cos{x} = 1. This is mathematics not physics, because this L has no physical meaning.

The physics in the action integral comes from (and only from) the correct functional dependence of the Lagrangian on the dynamical variables.

Didn't you learn from elementary mechanics that

L = T - V

Notice this; While T = 0 in your "Lagrangian", your "solutions" (on the other hand) have non-vanishing T! OK, I leave you to figure out your mistake!

Look, we call

L_{1} = T = \frac{1}{2} \left( \dot{x}^{2} + \dot{y}^{2} \right)

free system,

<br /> L_{2} = \frac{1}{2} \left( \dot{x}^{2} + \dot{y}^{2} - x^{2} - y^{2} \right)<br />

harmonic oscillator, and

L_{3} = L_{2} + \frac{1}{2} \left( y \dot{x} - x \dot{y} \right)

forced oscillator. So what you called "Lagrangian" is not a Lagrangian. It is nothing but the generalized POTENTIAL of the forced oscillator L_{3};

<br /> V(x,y, \dot{x}, \dot{y}) = \frac{1}{2} \left( x \dot{y} - y \dot{x} + x^{2} + y^{2} \right)<br />

To derive the dynamical equations from this potential, one uses

\ddot{x} = - \frac{\partial V}{\partial x} + \frac{d}{dt} \left( \frac{\partial V}{\partial \dot{x}} \right)

and a similar equation for y. Notice that the same equations of motion follow from the E-L equations when applied to L_{3}


sam

 

[jostpuur]

Didn't you learn from elementary mechanics that

L = T - V

I learned this in mechanics. When I encountered the Dirac field's Lagrangian, I noticed that not all Lagrangian's have this form, and came up with the simple the Lagrangian of my OP which had similar features with the Dirac field's Lagrangian. I then learned, from pair of PF members, that these kind of Lagrange's functions are called non-regular, and that there are more concrete examples of these with magnetic fields, like explained here http://en.wikipedia.org/wiki/Dirac_bracket. The most surprising fact to me was that the Lagrangian of general relativity is also of this non-regular type.

IMO the quantization procedure you have explained seems very ad hoc. You just put the anti-commutations in by force. I hope there is more to the understanding of Fermi fields, than pure acceptance of the anti-commutating operators and Grassmann numbers.

 

[jostpuur]

I have difficulty with these Grassmann numbers in the classical context. If I have an operator A that corresponds to some physical quantity, then we usually interpret the expectation value of this as the corresponding classical quantity.

<br /> A_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle<br />

Now suppose there is another quantity, and an operator B for it, so that AB+BA=0. What is the product of these classical quantities? I would say it's this:

<br /> A_{\textrm{classical}}B_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle \langle \Psi|B|\Psi\rangle<br />

And the classical values commute normally. So indeed, how do the Grassmann numbers enter this business?

 

[Rainbow Child]

I then learned, from pair of PF members, that these kind of Lagrange's functions are called non-regular, and that there are more concrete examples of these with magnetic fields, like explained here http://en.wikipedia.org/wiki/Dirac_bracket. The most surprising fact to me was that the Lagrangian of general relativity is also of this non-regular type.

IMO the quantization procedure you have explained seems very ad hoc. You just put the anti-commutations in by force. I hope there is more to the understanding of Fermi fields, than pure acceptance of the anti-commutating operators and Grassmann numbers.

Holy cow! In this url is the example that I posted! .

 

[Rainbow Child]

I have difficulty with these Grassmann numbers in the classical context. If I have an operator A that corresponds to some physical quantity, then we usually interpret the expectation value of this as the corresponding classical quantity.

<br /> A_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle<br />

Now suppose there is another quantity, and an operator B for it, so that AB+BA=0. What is the product of these classical quantities? I would say it's this:

<br /> A_{\textrm{classical}}B_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle \langle \Psi|B|\Psi\rangle<br />

So indeed, how do the Grassmann numbers enter this business?

There is no classical physical analogue for Grassmann numbers

 

[jostpuur]

There is no classical physical analogue for Grassmann numbers

Yeah well... but the reason why they are said to be not physical, is because they are Grassmann numbers, and now I'm wondering that why are there these Grassmann numbers. People say they come on the limit \hbar\to 0 out of the operators, but I'm not so convinced.

 

[reilly]

Thanks

I had to spend a few days deciding whether or not I'm serious. My kids, all in their 40s used to think I was silly, but now they think I'm a serious guy. And, after many consultations with academic colleagues, my students and business partners, I conclude that indeed I'm serious -- why, even my jazz musician buddies think I'm a serious cat..

Now, serious or not, I'm kind of simple minded soul.When I see "quantization of", I'm afraid that I think of q's and p's, creation and destruction operators and stuff like that. So, I made an error, which jostspuur discussed in a kind way. Big friggin' deal.

I congratulate you on your post on Grassmann variables.

And of course, I will be grateful for years to come for your advice, which I will refrain from calling gratuitous.
Regards,
Reilly Atkinson

Being a bit gun-shy, I forgot to mention that this type of problem is discussed in detail, ignorable variables, in Lanczos. Even without Grassmann, it's tricky stuff, but well worth reading.

ARE YOU SERIOUS? Before posting your "answer" try to understand the question raised by the OP.
The man asked about the peculiar (classical & quantum mechanical) nature of Fermi fields.

regards

sam

 

[jostpuur]

<br /> p_x = y<br />
<br /> p_y = -x<br />

<br /> H=x^2 + y^2<br />

So the Wikipedia's page, http://en.wikipedia.org/wiki/Dirac_bracket, explains, that I should take a new Hamilton's function

<br /> H=x^2 + y^2 + u(p_x - y) + v(p_y + x)<br />

with some arbitrary smooth functions u(x,y,p_x,p_y) and v(x,y,p_x,p_y). The equations of motion now become

<br /> \dot{x} = \frac{\partial H}{\partial p_x} = u<br />
<br /> \dot{y} = \frac{\partial H}{\partial p_y} = v<br />
<br /> \dot{p}_x = -\frac{\partial H}{\partial x} = -2x - v<br />
<br /> \dot{p}_y = -\frac{\partial H}{\partial y} = -2y + u<br />

The functions u and v can be eliminated, and we get

<br /> \dot{p}_x = -2x - \dot{y}<br />
<br /> \dot{p}_y = -2y + \dot{x}<br />

Finally, by substituting \dot{p}_x-\dot{y}=0 and \dot{p}_y+\dot{x}=0, we get the same equations of motion

<br /> \dot{x} = y<br />
<br /> \dot{y} = -x<br />

that were also implied by the original Lagrange's function. All this seems to make sense, but I have difficulty understanding how the quantization happens. Where is the Dirac's bracket coming from? Why not quantize the system by writing down the Schrödinger's equation

<br /> i\hbar\partial_t \Psi = \big(x^2 + y^2 - uy + vx - i\hbar(u\partial_x + v\partial_y)\big)\Psi?<br />

 

[jostpuur]

I also tried to quantize this by regularizing the system like this

<br /> L = \epsilon(\dot{x}^2 + \dot{y}^2) + \dot{x}y - x\dot{y} - x^2 - y^2<br />

The canonical momenta are

<br /> p_x = 2\epsilon\dot{x} + y<br />
<br /> p_y = 2\epsilon\dot{y} - x<br />

and the system can be represented with a Hamilton's function

<br /> H=\frac{1}{4\epsilon}(p_x^2 \;+\; p_y^2) \;+\; \frac{1}{2\epsilon}(xp_y \;-\; yp_x) \;+\; \big(1\;+\;\frac{1}{4\epsilon}\big)(x^2 \;+\; y^2)<br />

as usual. If one goes through the labor of finding solutions to this, it becomes evident that all solutions do not converge at the limit \epsilon\to 0. However, the energies of these solutions diverge too. This means, that this system approximates the original system in the following sense: No matter how small the epsilon is, there always exists some high energy solutions that are far different from the solutions of the original system. On the other hand, if we restrict the attention to low energy solutions only, then by setting epsilon sufficiently small, the solutions become approximately solutions of the original system.

In polar coordinates the Schrödinger's equation becomes

<br /> i\partial_t\Psi(t,r,\theta) = \Big(-\frac{1}{4\epsilon}\big(\partial_r^2 \;+\; \frac{1}{r}\partial_r \;+\; \frac{1}{r^2}\partial_{\theta}^2\big) \;-\; \frac{i}{2\epsilon}\partial_{\theta} \;+\; \big(1 \;+\; \frac{1}{4\epsilon}\big)r^2\Big)\Psi(t,r,\theta)<br />

I think I've succeeded in solving the energy eigenstates of this. Firstly

<br /> \psi(r,\theta) = \exp\big(-\frac{1}{2}\sqrt{1+4\epsilon} r^2\big)<br />

is a solution. In analogy with the harmonic oscillator, it makes sense to next attempt solutions of form

<br /> \psi(r,\theta) = H(r,\theta) \exp\big(-\frac{1}{2}\sqrt{1+4\epsilon} r^2\big),<br />

and substitute the attempt

<br /> H(r,\theta) = \sum_{k_1=0}^{\infty} \sum_{k_2=-\infty}^{\infty} a_{k_1,k_2} r^{k_1} e^{ik_2\theta}.<br />

I don't want to go into details now, but it turns out that one can get needed recursion relations for the coefficients a_{k_1,k_2}. The energy spectrum becomes

<br /> E_{n,k} = \frac{\sqrt{1+4\epsilon}(|k| + 2n + 1) + k}{2\epsilon},\quad n\in\{0,1,2,\ldots\},\;k\in\mathbb{Z}<br />

What I found somewhat surprising, is that all energies diverge towards infinity when one sets \epsilon\to 0, but in fact it is possible to make sense out of this intuitively. I thought about it like this: Classically, with epsilon set to zero, all finite energy solutions are forced on the circular paths. In some sense, if particle was forced out of the circular path, then its energy would go to infinity. Then suppose we have a quantum mechanical wave packet on the circular path. The wave packet cannot go sufficiently accurately on the circular path, and always has some amplitudes for being outside the circular path. Hence the infinite energies.

But why did I still not encounter anything the resembles fermions? Nothing is anti-commuting here. Is this simply a wrong way to attempt to quantize the system? If you consider the strong magnetic field approximation, the IMO, this would seem very well justified way to quantize the system.

 

[strangerep]

The equations of motion now become
<br /> \dot{x} = \frac{\partial H}{\partial p_x} = u<br />
<br /> \dot{y} = \frac{\partial H}{\partial p_y} = v<br />
<br /> \dot{p}_x = -\frac{\partial H}{\partial x} = -2x - v<br />
<br /> \dot{p}_y = -\frac{\partial H}{\partial y} = -2y + u<br />

I don't think you're allowed to write down those eqns of motion in a
singular (constrained) system. The Wiki page you quoted says that
the eqn of motion is
<br /> \Big(\partial_q H + \dot p \Big)\delta q + \Big(\partial_p H - \dot q \Big)\delta p ~\approx~ 0<br />
where "\approx" means "weak equality". You're not allowed to set the
\delta q,~\delta p to zero separately (to get the usual eqns of motions)
because the variations are restricted by a constraint.

Where is the Dirac's bracket coming from?

If you carry through the constrained quantization procedure further, and more
carefully, (as explained on the Wiki page), I think you'll find that the usual
Poisson bracket is modified (in general) by the presence of constraints.

Why not quantize the system by writing down the Schrödinger's equation
<br /> i\hbar\partial_t \Psi = \big(x^2 + y^2 - uy + vx - i\hbar(u\partial_x + v\partial_y)\big)\Psi?<br />

You haven't yet constructed a representation of the various classical observables
(functions on phase space) as unitary operators on a Hilbert space.

For fermions, you need to start from a classical phase space based on
Grassmann variables.

P.S. Dirac's little booklet "Lectures on Quantum Mechanics" explains this stuff
far more pedagogically than the Wiki page. I picked up a copy from Amazon
quite cheaply.

 

[jostpuur]

For fermions, you need to start from a classical phase space based on
Grassmann variables.

I keep hearing this all the time, but I'm not convinced. I have two remarks.

My post #31: The anti-commutation of some operators does not imply anti-commutation of the classical variables.

Rainbow Child's post #11: He explains, that the we should be getting anti-commuting operators, somehow with the Dirac's brackets, even though he did not mention classical Grassmann variables anywhere.

It is difficult for me to tell if this classical Grassmann variable thing is a myth or fact.

P.S. Dirac's little booklet "Lectures on Quantum Mechanics" explains this stuff
far more pedagogically than the Wiki page. I picked up a copy from Amazon
quite cheaply.

I hope you are talking about this

https://www.amazon.com/dp/0486417131/?tag=pfamazon01-20

It should be coming towards me in the post soon.

 

[strangerep]

In post #31:

If I have an operator A that corresponds to some physical quantity, then we usually interpret the expectation value of this as the corresponding classical quantity.
<br /> A_{cl} = \langle \Psi|A|\Psi\rangle<br />

Now suppose there is another quantity, and an operator B for it, so that AB+BA=0.
What is the product of these classical quantities? I would say it's this:

<br /> A_{cl}B_{cl} = \langle \Psi|A|\Psi\rangle \langle \Psi|B|\Psi\rangle<br />

For fermionic A,B the above can't be right.
Consider the case B=A: It could well be the case that \langle A\rangle \ne 0,
but we'll always find \langle A^2\rangle = 0 .

Rainbow Child's post #11: He explains, that the we should be getting anti-commuting operators, somehow with the Dirac's brackets, even though he did not mention classical Grassmann variables anywhere.

In post #11, I didn't see anything about anti-commuting operators. The Dirac brackets
that R.C. mentioned are defined on the Wiki page quoted earlier. They are generalizations
of Poisson brackets. No anti-commutation is involved, since it's dealing with
Poisson brackets of real/complex-valued functions on phase space.

It's crucial to distinguish the notions of (1) constraints and Dirac brackets, and (2)
fermionic anti-commutators. The Wiki page mostly deals with constrained classical
(and bosonic) systems. For example, the M_{ab} mentioned therein is
composed from the 2nd-class constraints (i.e: Poisson brackets of constraints that
don't commute with other constraints). For classical and bosonic systems, there's
always an even number of these, hence a square matrix makes sense. Normally,
a PB of something with itself is zero, but for a Grassmann-valued field F it's
possible that a Poisson bracket of F with itself won't vanish. Hence the method
must be adjusted accordingly (but the Wiki page doesn't elaborate on this).

The Wiki page mentions that constraints are always "applicable" to fermions
because the Lagrangian is linear in the velocities for fermions (think of the
Lagrangian for a free Dirac electron).

It is difficult for me to tell if this classical Grassmann variable thing is a myth or fact.

I'm not sure what you mean by "myth" and "fact" here. It is certainly a fact that
path integral methods in QFT for fermions use Grassmann variables. Whether this
is just an ad-hoc mathematical artifice, or indicative of some deeper physical truth,
depends on one's philosophy.

I hope you are talking about this
https://www.amazon.com/dp/0486417131/?tag=pfamazon01-20

Yes. Note that it talks mainly about constraints, and the techniques to deal
with them. Not much about about fermions specifically.

If you have a copy of Peskin & Schroeder, you can find a little bit about Grassmann
fields and the associated calculus in section 9.5. Also eq(9.75) for Grassmann
derivatives, but you'll probably need to find another source with a more extensive
treatment of the latter. You'll need Grassmann derivatives to see how classical
Poisson brackets become anti-commuting in the case of Grassmann fields.
The Wiki page alludes to this, but only very briefly. A (much) more advanced
treatment is in Henneaux & Teitelboim

 

[jostpuur]

I should probably wait until I get the Dirac's lectures. This is getting a little bit speculative from me now... but you know, it's so difficult to stop thinking!

What happens, if you do not assign any Grassmann properties to the classical field variables, but still quantize the field by starting from the Dirac field Lagrangian. Is the canonical quantization still going to give anti-commuting field operators?

My problem is this, at the moment: As long as I don't know what happens with canonical quantization, starting with the commuting classical fields, and with the Dirac's Lagrangian, I keep hoping that this will give the correct anti-commuting quantum field.

I just made an interesting remark. The Peskin & Schroeder don't mention a thing about the Grassmann variables when they first talk about the Dirac's field, but later, with path integral quantization, they introduce the Grassmann variables. Could it be, that the Grassmann variables are intended to be used precisely with the path integral quantization, and not with the canonical operator quantization?

 

[jostpuur]

It could be I got the regularization attempt to its end. The eigenstates and -energies of the quantized system, described by the Lagrangian

<br /> L=\dot{x}y-x\dot{y}-x^2-y^2<br />

should be

<br /> \psi_n(r,\theta) = r^n e^{-\frac{1}{2}r^2-in\theta},\quad E_n=E_{\textrm{zero}} + n,\quad n\in\{0,1,2,3,\ldots\},<br />

assuming I did everything right. These wave functions and energies were obtained by taking limit \epsilon\to 0 of the solutions of the regularized system. The zero point energy diverged towards infinity, but some of the energy differences remained finite. I'm not sure what this all means. Or is this nonsense? Some math it is, at least.

 

[genneth]

Like Rainbow Child said, this particular problem runs into difficulties due to its classical structure. The non-regularity is a fairly serious problem. Another view is to remind yourself that quantisation is not a procedure. It is at best a heuristic for solving the inverse to taking the classical limit. Various people here have suggested ways to "quantise" the system, but not all of them will contain the relevant physics that you are looking for, even if they all formally reduce down to the same equations of motion.

 

[strangerep]

What happens, if you do not assign any Grassmann properties to the classical field variables, but still quantize the field by starting from the Dirac field Lagrangian. Is the canonical quantization still going to give anti-commuting field operators?

I'll work it out in detail later.

Peskin & Schroeder don't mention a thing about the Grassmann variables when they first talk about the Dirac's field, but later, with path integral quantization, they introduce the Grassmann variables. Could it be, that the Grassmann variables are intended to be used precisely with the path integral quantization, and not with the canonical operator quantization?

Look at the last paragraphs on p56, before eq(3.96). P&S impose anti-commutation
relations arbitrarily (after going through the usual arguments about how commutators
don't work for fermions). Then again on p58 near eqs(3.101, 3.102). They are
imposing anti-commutation arbitrarily.

So either way, be it with canonical quantization or path integrals, they're
putting in anti-commutation by hand.

 

[strangerep]

Jostpuur,

Back in your posts #36 and #37, I don't think you were
applying the Dirac-Bergman constraint quantization method
incorrectly. Here's my attempt at it...

Starting from the Lagrangian
<br /> L ~=~ \dot x y - z\dot y - x^2 - y^2 ~,<br />
the canonical momenta are
<br /> p_x ~=~ \frac{\partial L}{\partial \dot x} ~=~ y ~;~~~~<br /> p_y ~=~ \frac{\partial L}{\partial \dot y} ~=~ -x<br />
and the "standard" Hamiltonian is
<br /> H ~=~ p_x \dot x + p_y \dot y - L ~=~ x^2 + y^2 ~=:~ V(x,y)<br />
Following the Dirac-Bergman method, we have the constraint functions:
<br /> \phi_1 := p_x - y ~,~~~~ \phi_2 := p_y + x<br />
The standard Poisson bracket is defined by
<br /> \{F,G\} ~=~ \sum_i \left(<br /> \frac{\partial F}{\partial q_i} \frac{\partial G}{\partial p_i}<br /> - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i}<br /> \right) ~~~~~~(PB)<br />
So the only non-vanishing Poisson bracket between the constraint
functions is
<br /> \{\phi_1, \phi_2\}<br /> ~=~ \partial_y \phi_1 \partial_{p_y} \phi_2<br /> - \partial_{p_x} \phi_1 \partial_x \phi_2<br /> ~=~ -2 ~.<br />
Therefore, \phi_1 and \phi_2 are
"2nd-class constraints".

To get the Dirac bracket, we need the matrix
M_{ab} := \{\phi_a, \phi_b\} and its inverse...
which here is
<br /> [M_{ab}] ~:=~<br /> \left(\begin{array}{cc} 0 &amp; -2 \\ 2 &amp; 0 \end{array}\right)<br /> ~;~~~~~~<br /> [M_{ab}]^{-1} ~:=~<br /> \frac{1}{2}\left(\begin{array}{cc} 0&amp;1\\ -1&amp;0 \end{array}\right)<br />
and the standard Dirac bracket is then given by
<br /> \{F,G\}_{DB} ~=~ \{F,G\}<br /> ~-~ \sum_{ab} \{F,\phi_a\}~M^{-1}_{ab}~\{\phi_b,G\}<br /> ~~~~~~(DB1)<br />
or in our case,
<br /> \{F,G\}_{DB} ~=~ \{F,G\}<br /> ~-~ \frac{1}{2} ~ \left(\{F,\phi_1\},~ \{F,\phi_2\}\right)<br /> \left(\begin{array}{cc} 0&amp;1\\ -1&amp;0 \end{array}\right)<br /> {\{\phi_1,G\} \choose \{\phi_2,G\}}<br /> ~~~~~~(DB2)<br />
Writing out some ordinary Poisson brackets between
x,y,p_x,p_y and the constraints, we find only the
following are non-zero:
<br /> \{x,\phi_1\} = \{y,\phi_2\} = 1 ~;~~~~<br /> \{p_x,\phi_2\} = -1 ~;~~~~ \{p_y,\phi_1\} = 1 ~.<br />
Poisson brackets between x,y,p_x,p_y are
<br /> \{x,y\} = \{p_x,p_y\} = \{y,p_x\} = \{x,p_y\} = 0 ~;~~~~<br /> \{x,p_x\} = \{y,p_y\} = 1 ~;~~~~<br />

Now we can compute the Dirac brackets. I find that only the following
ones are non-zero:
<br /> \{x,y\}_{DB} = \{x,p_x\}_{DB}<br /> = \{y,p_y\}_{DB} = \{p_x,p_y\}_{DB} = \frac{1}{2}<br />
By the standard prescription, we can quantize the theory by using
the original Hamiltonian \hat H := \hat x^2 + \hat y^2,
together with the commutation relations:
<br /> [\hat x,\hat y] = [\hat x,\hat p_x]<br /> = [\hat y,\hat p_y] = [\hat p_x,\hat p_y]<br /> = \frac{i\hbar}{2} ~~~~~~(CR)<br />
This is very similar to earlier posts by Rainbow Child and
samalkhaiat (except for some slight differences in signs and
factors).
[Continued in next post...]

 

[strangerep]

Jostpuur,

Back in your posts #36 and #37, I think you were
applying the Dirac-Bergman constraint quantization
method incorrectly. Here's my attempt at it...

Starting from the Lagrangian
<br /> L ~=~ \dot x y - z\dot y - x^2 - y^2 ~,<br />
the canonical momenta are
<br /> p_x ~=~ \frac{\partial L}{\partial \dot x} ~=~ y ~;~~~~<br /> p_y ~=~ \frac{\partial L}{\partial \dot y} ~=~ -x<br />
and the "standard" Hamiltonian is
<br /> H ~=~ p_x \dot x + p_y \dot y - L ~=~ x^2 + y^2 ~=:~ V(x,y)<br />
Following the Dirac-Bergman method, we have the constraint functions:
<br /> \phi_1 := p_x - y ~,~~~~ \phi_2 := p_y + x<br />
The standard Poisson bracket is defined by
<br /> \{F,G\} ~=~ \sum_i \left(<br /> \frac{\partial F}{\partial q_i} \frac{\partial G}{\partial p_i}<br /> - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i}<br /> \right) ~~~~~~(PB)<br />
So the only non-vanishing Poisson bracket between the constraint
functions is
<br /> \{\phi_1, \phi_2\}<br /> ~=~ \partial_y \phi_1 \partial_{p_y} \phi_2<br /> - \partial_{p_x} \phi_1 \partial_x \phi_2<br /> ~=~ -2 ~.<br />
Therefore, \phi_1 and \phi_2 are
"2nd-class constraints".

To get the Dirac bracket, we need the matrix
M_{ab} := \{\phi_a, \phi_b\} and its inverse...
which here is
<br /> [M_{ab}] ~:=~<br /> \left(\begin{array}{cc} 0 &amp; -2 \\ 2 &amp; 0 \end{array}\right)<br /> ~;~~~~~~<br /> [M_{ab}]^{-1} ~:=~<br /> \frac{1}{2}\left(\begin{array}{cc} 0&amp;1\\ -1&amp;0 \end{array}\right)<br />
and the standard Dirac bracket is then given by
<br /> \{F,G\}_{DB} ~=~ \{F,G\}<br /> ~-~ \sum_{ab} \{F,\phi_a\}~M^{-1}_{ab}~\{\phi_b,G\}<br /> ~~~~~~(DB1)<br />
or in our case,
<br /> \{F,G\}_{DB} ~=~ \{F,G\}<br /> ~-~ \frac{1}{2} ~ \left(\{F,\phi_1\},~ \{F,\phi_2\}\right)<br /> \left(\begin{array}{cc} 0&amp;1\\ -1&amp;0 \end{array}\right)<br /> {\{\phi_1,G\} \choose \{\phi_2,G\}}<br /> ~~~~~~(DB2)<br />
Writing out some ordinary Poisson brackets between
x,y,p_x,p_y and the constraints, we find only the
following are non-zero:
<br /> \{x,\phi_1\} = \{y,\phi_2\} = 1 ~;~~~~<br /> \{p_x,\phi_2\} = -1 ~;~~~~ \{p_y,\phi_1\} = 1 ~.<br />
Poisson brackets between x,y,p_x,p_y are
<br /> \{x,y\} = \{p_x,p_y\} = \{y,p_x\} = \{x,p_y\} = 0 ~;~~~~<br /> \{x,p_x\} = \{y,p_y\} = 1 ~;~~~~<br />
[Continued in next post...]

 

[strangerep]

(Continuation of post #46...)

Now we can compute the Dirac brackets. I find that only the following
ones are non-zero:
<br /> \{x,y\}_{DB} = \{x,p_x\}_{DB}<br /> = \{y,p_y\}_{DB} = \{p_x,p_y\}_{DB} = \frac{1}{2}<br />
By the standard prescription, we can quantize the theory by using
the original Hamiltonian \widehat H := \hat x^2 + \hat y^2,
together with the commutation relations:
<br /> [\hat x,\hat y] = [\hat x,\hat p_x]<br /> = [\hat y,\hat p_y] = [\hat p_x,\hat p_y]<br /> = \frac{i\hbar}{2} ~~~~~~(CR)<br />
This is very similar to earlier posts by Rainbow Child and
samalkhaiat (except for some slight differences in signs and
factors).

I.e., the time-evolution of any observable operator
\hat A = \hat A(\hat x, \hat y) in the Hilbert space
for this theory is given by
<br /> i\hbar ~ \frac{\partial \hat A}{\partial t} ~=~ [\hat A, \hat H] ~.<br />

But all this still bosonic. There's no way to somehow turn the handle
further and extract anti-commutators. (I think you were misinterpreting
RC's remarks about using the Dirac bracket to "get" anti-commutators.
I don't think that's what RC actually meant.) You can't get anti-commutators
from commutators.

Instead, you've got to take the approach that samalkhaiat explained
earlier, and use Grassmann numbers. But then, (as he also explained),
terms like x^2 in your Lagrangian are identically zero.
Game over. The Lagrangian cannot represent a "fermionic oscillator".
But neither can it represent a physical bosonic system because it lacks
quadratic momentum (kinetic energy) terms. (This is what samalkhaiat
was trying to emphasize earlier).

 

[jostpuur]

I'm so out of time at the moment! I'm forced to delay thinking about those things later.

I'll say one comment on this

Look at the last paragraphs on p56, before eq(3.96). P&S impose anti-commutation
relations arbitrarily (after going through the usual arguments about how commutators
don't work for fermions). Then again on p58 near eqs(3.101, 3.102). They are
imposing anti-commutation arbitrarily.

So either way, be it with canonical quantization or path integrals, they're
putting in anti-commutation by hand.

(because I had thought about this earlier already.) In the chapter Dirac Field, P&S put anti-commutation relations to the fields only in the quantization. They start the chapter with the classical Dirac field, and there is no mentioning about anti-commuting Grassmann numbers in that context yet. They let the reader assume that the classical Dirac field is \psi\in\mathbb{C}^4, and it is only the quantized field that anti-commutes. I'm still trying to keep hopes up for the possibility, that the anti-commuting classical variables would belong only to the path integral quantization, because at the moment it seems the only way this could start making sense.

I used to call the P&S Introduction to the QFT a "bible of QFT", because the proofs are left as matter of "faith". Now when I'm trying to see where the anti-commuting numbers really belong to, I can see that it is also possible to interpret this book in different ways!

 

[jostpuur]

I'll be making progress with this slowly but firmly.

Jostpuur,

Back in your posts #36 and #37, I think you were
applying the Dirac-Bergman constraint quantization
method incorrectly.

I see the #36 was on completely different track than the Dirac-Bergman method. Although my calculation is probably not total nonsense, because the equations of motion were right in the end... or then it was lucky nonsense?

The #37 was not supposed to be Dirac-Bergman quantization. It was my own regularization attempt. Now I'm keeping hopes up, that the same energy spectrum would also follow with the constraint approach, because then there would be a chance that the regularization was not nonsense.

Here's my attempt at it...

Starting from the Lagrangian
<br /> L ~=~ \dot x y - z\dot y - x^2 - y^2 ~,<br />
the canonical momenta are
<br /> p_x ~=~ \frac{\partial L}{\partial \dot x} ~=~ y ~;~~~~<br /> p_y ~=~ \frac{\partial L}{\partial \dot y} ~=~ -x<br />
and the "standard" Hamiltonian is
<br /> H ~=~ p_x \dot x + p_y \dot y - L ~=~ x^2 + y^2 ~=:~ V(x,y)<br />
Following the Dirac-Bergman method, we have the constraint functions:
<br /> \phi_1 := p_x - y ~,~~~~ \phi_2 := p_y + x<br />

I see, and the equations of motion are

<br /> \dot{x}\; =\; \frac{\partial H}{\partial p_x}\; +\; u_1\frac{\partial\phi_1}{\partial p_x}\;+\; u_2 \frac{\partial\phi_2}{\partial p_x} \; =\; u_1,<br />

<br /> \dot{y}\; =\; \frac{\partial H}{\partial p_y}\; +\; u_1\frac{\partial\phi_1}{\partial p_y}\;+\; u_2 \frac{\partial\phi_2}{\partial p_y} \; =\; u_2,<br />

<br /> \dot{p}_x\; =\; -\frac{\partial H}{\partial x}\; -\; u_1\frac{\partial\phi_1}{\partial x}\;-\; u_2 \frac{\partial\phi_2}{\partial x} \; =\; -2x -u_2,<br />

<br /> \dot{p}_y\; =\; -\frac{\partial H}{\partial y}\; -\; u_1\frac{\partial\phi_1}{\partial y}\;-\; u_2 \frac{\partial\phi_2}{\partial y} \; =\; -2y +u_1.<br />

The standard Poisson bracket is defined by
<br /> \{F,G\} ~=~ \sum_i \left(<br /> \frac{\partial F}{\partial q_i} \frac{\partial G}{\partial p_i}<br /> - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i}<br /> \right) ~~~~~~(PB)<br />
So the only non-vanishing Poisson bracket between the constraint
functions is
<br /> \{\phi_1, \phi_2\}<br /> ~=~ \partial_y \phi_1 \partial_{p_y} \phi_2<br /> - \partial_{p_x} \phi_1 \partial_x \phi_2<br /> ~=~ -2 ~.<br />
Therefore, \phi_1 and \phi_2 are
"2nd-class constraints".

I don't understand how one could see from this what are second class constraints.

If g(x,y,p_x,p_y) is some function of the coordinates, it has the equation of motion

<br /> \dot{g}\approx [g,H+u_m\phi_m]<br />

Right now there is

<br /> \frac{\partial}{\partial x}(u_m\phi_m) \approx u_2,\quad \frac{\partial}{\partial y}(u_m\phi_m) \approx -u_1,\quad \frac{\partial}{\partial p_x}(u_m\phi_m)\approx u_1,\quad \frac{\partial}{\partial p_y}(u_m\phi_m)\approx u_2<br />

so

<br /> \dot{g}\approx \frac{\partial g}{\partial x} u_1 + \frac{\partial g}{\partial y} u_2 - \frac{\partial g}{\partial p_x}(u_2 + 2x) + \frac{\partial g}{\partial p_y}(u_1-2y)<br />

If I substitute g=\phi_1 and g=\phi_2, I get

<br /> \dot{\phi}_1 \approx 2(u_2+x)<br />

<br /> \dot{\phi}_2 \approx 2(u_1-y).<br />

Am I now correct to say that

<br /> p_x - y = 0,\quad p_y + x = 0<br />

are the primary constraints, and

<br /> u_2+x = 0,\quad u_1-y = 0<br />

are the secondary constraints?

About 2/3 of math in #46 & #47 still ahead...

 

[jostpuur]

Oh, how confusing. Only now I noticed that the EOM in #36 and #49 were exactly the same! Well, I derived them from completely different starting point, at least. I'm not sure if that was coincidental... In #36 I merely calculated the usual Hamilton's equations, starting with the modified H, and applied the constraint condition in the end. In #49 I started with the equations with the Lagrange's multipliers, in the way it was supposed to be done.

The SE in the bottom of #36 was nonsense at least.