Quantizing a two-dimensional Fermion Oscillator

In summary, the problem with quantizing a two dimensional system defined by the Lagrange's function is that the equations of motion are non-trivial and the velocities are not independent of the coordinates. The attempt with the Schrödinger's equation and Hamiltonian fails because the canonical momenta are y and x. The path integrals and action cannot be used to define time evolution because there are no terms in the S that are proportional to \frac{1}{t_1-t_0}.
  • #1
jostpuur
2,116
19
Problem:

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

[tex]
L=\dot{x}y - x\dot{y} - x^2 - y^2 ?
[/tex]

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

[tex]
\dot{x}(t) = y(t)
[/tex]
[tex]
\dot{y}(t) = -x(t),
[/tex]

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

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

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

[tex]
L=\textrm{Im}(\dot{z}^* z) - |z|^2
[/tex]
[tex]
\dot{z} = -iz
[/tex]
[tex]
z(t)=e^{-it}z(0)
[/tex]

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

[tex]
H=\frac{\partial L}{\partial\dot{x}}\dot{x} + \frac{\partial L}{\partial\dot{y}}\dot{y} - L = x^2 + y^2 = |z|^2
[/tex]

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

[tex]
i\partial_t\Psi(t,z) = |z|^2\Psi(t,z),
[/tex]

and solutions are

[tex]
\Psi(t,z) = e^{-i|z|^2 t}f(z).
[/tex]

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

[tex]
p_x = \frac{\partial L}{\partial\dot{x}} = y
\quad\quad\quad
p_y = \frac{\partial L}{\partial\dot{y}} = -x
[/tex]

so when we should leave parameters x and y untouched, and substitute [itex]p_x\to -i\partial_x[/itex] and [itex]p_y\to -i\partial_y[/itex], 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 [itex]t_1-t_0[/itex]. We can parametrize a path

[tex]
z(t) = \frac{(t_1-t)z' + (t-t_0)z}{t_1-t_0},
[/tex]

and compute the action

[tex]
S = \int\limits_{t_0}^{t_1} L(z(t))dt = xy' - yx' - \frac{1}{3}(t_1-t_0)\big( x^2 + xx' + (x')^2 + y^2 + yy' + (y')^2\big),
[/tex]

but this cannot be used to define time evolution with

[tex]
\Psi(t+\Delta t, z) = N \int dx'\;dy'\; e^{iS}\Psi(t,z')\;+\; O(\Delta t^2)
[/tex]

as usual, because there is no

[tex]
\propto \quad\frac{1}{t_1-t_0},\quad\quad\textrm{or}\quad \frac{1}{(t_1-t_0)^2}
[/tex]

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 [itex]\psi(0,x)[/itex] 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

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

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

[tex]
\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)
[/tex]

So the fermion analogy to the harmonic oscillator is

[tex]
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
[/tex]

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 [itex]a^{\dagger}[/itex] and [itex]a[/itex] of the harmonic oscillator, these operators also have very explicit expressions in terms of the operators [itex]x[/itex] and [itex]-i\partial_x[/itex]. I would like to have something similarly explicit for the fermion oscillator too.
 
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  • #2
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... :frown:

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.
 
  • #3
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
 
  • #4
jostpuur said:
...
[tex]
H=\frac{\partial L}{\partial\dot{x}}\dot{x} + \frac{\partial L}{\partial\dot{y}}\dot{y} - L = x^2 + y^2 = |z|^2
[/tex]

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...
 
  • #5
The Lagrangian you posted is non-regular. In order for someone to pass from the Lagrangian description to the Hamiltonian one it must hold

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

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

[tex]p_i=\frac{\partial \,\mathcal{L}}{\partial\dot{q}^i}[/tex]

for the velocities, so the imply conditions on some of the [itex](p^i,q^j)[/itex], 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 [itex](x,y)[/itex] does not commute, i.e.

[tex]\{x,y\}_{Dirac}=-2 \quad (!)[/tex]

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 [itex] (difficulty\rightarrow \infty)[/itex]

Henneaux, Marc and Teitelboim, Claudio, Quantization of Gauge Systems. Princeton University Press, (1992)
 
  • #6
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
 
  • #7
reilly said:
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?

olgranpappy said:
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

[tex]
L = \dot{x}y - x\dot{y} + x^2 + y^2,
[/tex]

the Hamiltonian would be

[tex]
H=-|z|^2,
[/tex]

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

[tex]
z(t) = e^{it}z(0)
[/tex]

This system doesn't work very intuitively.

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

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

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

[tex]p_i=\frac{\partial \,\mathcal{L}}{\partial\dot{q}^i}[/tex]

for the velocities, so the imply conditions on some of the [itex](p^i,q^j)[/itex], 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 [itex](x,y)[/itex] does not commute, i.e.

[tex]\{x,y\}_{Dirac}=-2 \quad (!)[/tex]

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 :frown:

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?
 
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  • #8
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.
 
  • #9
reilly said:
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.
 
  • #10
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

[tex]
H = \boldsymbol{x}\times\boldsymbol{p} + |\boldsymbol{x}|^2?
[/tex]

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

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

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

[tex]
L = \dot{x}y - x\dot{y} - x^2 - y^2
[/tex]

[tex]
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
[/tex]

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

[tex]
H = xp_y - yp_x + x^2 + y^2
[/tex]
 
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  • #11
jostpuur said:
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

[tex]\{x,y\}_{Dirac}=-\frac{1}{2},\,\{x,p_x\}_{Dirac}=\{y,p_y\}_{Dirac}=\frac{1}{2}[/tex]

thus

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

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.
 
  • #12
Rainbow Child said:
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?
 
  • #13
jostpuur said:
("Lagrangian of GR is"?)

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

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! :smile:
 
  • #14
Hehe. Sorry for unclarity. I intended to point out your probable typos

the Lagrangian or GR in non-regular
:wink:


Yes! :smile:

Good piece of information to know.
 
  • #15
jostpuur said:
How do you quantize a two dimensional system defined by the Lagrange's function

[tex]
L=\dot{x}y - x\dot{y} - x^2 - y^2 ?
[/tex]

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

[tex]
\dot{x}(t) = y(t)
[/tex]
[tex]
\dot{y}(t) = -x(t),
[/tex]

This system represents a 2-dimensional harmonic oscillator;

[tex]
\ddot{x} + x = 0
[/tex]
[tex]
\ddot{y} + y = 0
[/tex]

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

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 [itex]a(p)[/itex]and [itex]a^{\dagger}(p)[/itex]. 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 [itex]\hbar \rightarrow 0[/itex] of an anticommutator in the quantum theory;

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

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 [itex]X_{i}(t)[/itex] and [itex]\bar{X}_{i}(t)[/itex] (i = 1,2,...,n) with the Lagrangian

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

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

[tex]\dot{X} = - i M^{-1} \frac{\partial V}{\partial \bar{X}}[/tex]

NOW COMES THE IMPORTANT BIT:

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

[tex]
H = V(X, \bar{X})
[/tex]
[tex]
\{ X_{i} , X_{j} \} = \{ X^{\dagger}_{i} , X^{\dagger}_{j} \} = 0
[/tex]
[tex]
\{ X_{i} , X^{\dagger}_{j} \} = M^{-1}_{ij}
[/tex]

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

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
 
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  • #16
reilly said:
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
 
  • #17
samalkhaiat said:
This system represents a 2-dimensional harmonic oscillator;

[tex]
\ddot{x} + x = 0
[/tex]
[tex]
\ddot{y} + y = 0
[/tex]

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

Wait a minute. We have

[tex]
\left\{\begin{array}{l}
\dot{x} = y\\
\dot{y} = -x\\
\end{array}\right.
\quad\implies\quad
\left\{\begin{array}{l}
\ddot{x} = -x\\
\ddot{y} = -y\\
\end{array}\right.
[/tex]

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.
 
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  • #18
Another idea: Could it be possible to quantize a system described by

[tex]
L=\epsilon(\dot{x}^2 + \dot{y}^2) + \dot{x}y - x\dot{y} - x^2 - y^2
[/tex]

as usual, and then take something reasonable out by setting [itex]\epsilon\to 0[/itex]?
 
  • #19
jostpuur said:
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 [itex]x^{2} + y^{2}[/itex]. 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 [itex]\dot{x}^{2}[/itex] in your Lagrangian. And to quantize a constrained bosonic system, follow the rule;

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

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

[tex]\{ \hat{A} , \hat{B} \}_{+} = i \hbar \{ A , B \}_{Dirac}[/tex]

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

[tex]
V = 0
[/tex]
[tex]
M_{ij} = \delta_{ij}
[/tex]
[tex]
\bar{X}_{i} = X_{i}
[/tex]

i.e.,

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

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

[tex] P_{i} + \frac{i}{2} \delta_{ij} X^{j} = 0[/tex]

the basic Dirac brackets are

[tex]\{ X_{i} , X_{j} \}_{Dirac} = - i \delta_{ij}[/tex]

and the quantization rule is

[tex]
\{ \hat{X}_{i} , \hat{X}_{j} \}_{+} = i \hbar \{ X_{i} , X_{j} \}_{Dirac} = \hbar \delta_{ij}
[/tex]

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
 
  • #20
samalkhaiat said:
... Also, it cannot represent a bosonic system because it is not quadratic in derivatives...

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

[tex]\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)[/tex]

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

[tex]\frac{q\,B_o}{m\,c}\gg 1[/tex]

we may neglect the kinetic term, arriving to

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

which is jostpuur's Lagrangian apart from numerical constants. :rolleyes:
 
  • #21
Rainbow Child said:
Actually we can make his system a physical one! :smile:
If we imagine a particle with charge [itex]q[/itex] and mass [itex]m[/itex] moving in the x-y plane along with a strong constant, homogeneous magnetic field towards the z-direction with strength [tex]B_o[/tex]. Let's also assume that there is an external potential [itex]V=\frac{m}{2}\,(x^2+y^2)[/itex].
Then the corresponding Lagragian reads

[tex]\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)[/tex]

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

[tex]\frac{q\,B_o}{m\,c}\gg 1[/tex]

we may neglect the kinetic term, arriving to

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

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

[tex]
\ddot{x} = \frac{q B_{0}}{2mc} \dot{y}
[/tex]
[tex]
\ddot{y} = - \frac{qB_{0}}{2mc} \dot{x}
[/tex]


regards


sam
 
  • #22
samalkhaiat said:
Rainbow Child said:
Why not neglect the potential term? This way you arrive at more sound dynamical equations

[tex]
\ddot{x} = \frac{q B_{0}}{2mc} \dot{y}
[/tex]
[tex]
\ddot{y} = - \frac{qB_{0}}{2mc} \dot{x}
[/tex]


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.
 
  • #23
samalkhaiat said:
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.

[tex]
L=\dot{x}y-x\dot{y}-x^2-y^2
[/tex]

[tex]
\left\{\begin{array}{l}
D_t\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}\\
D_t\frac{\partial L}{\partial \dot{y}} = \frac{\partial L}{\partial y}\\
\end{array}\right.
\quad\implies\quad \left\{\begin{array}{l}
\dot{y} = -\dot{y} - 2x\\
-\dot{x} = \dot{x} - 2y\\
\end{array}\right.
\quad\implies\quad \left\{\begin{array}{l}
\dot{x} = y\\
\dot{y} = -x\\
\end{array}\right.
[/tex]

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.
 
  • #24
Rainbow Child said:
samalkhaiat said:
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

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

or

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

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

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

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

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
 
  • #25
jostpuur said:
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

[tex]L = x \dot{y} - y \dot{x}[/tex]

with

[tex]xy + yx = 0[/tex]

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
 
Last edited:
  • #26
So if I ask a question "How do you quantize the system [itex]L=\dot{x}y-x\dot{y}-x^2-y^2[/itex]?", your answer is "No, that is wrong. You have to quantize something else."
 
  • #27
jostpuur said:
So if I ask a question "How do you quantize the system [itex]L=\dot{x}y-x\dot{y}-x^2-y^2[/itex]?", 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
 
  • #28
samalkhaiat said:
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

jostpuur said:
[tex]
L=\dot{x}y - x\dot{y} - x^2 - y^2 ?
[/tex]

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

[tex]
\dot{x}(t) = y(t)
[/tex]
[tex]
\dot{y}(t) = -x(t),
[/tex]

and whose solutions are

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

I don't understand why this would not be a well defined classical system.
 
  • #29
I hope this is not a waste of time!

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

[tex]L = t \dot{x} + \sin{x}[/tex]

leads to the E-L equation [itex] \cos{x} = 1[/itex]. 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

[tex] L = T - V[/tex]

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

Look, we call

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

free system,

[tex]
L_{2} = \frac{1}{2} \left( \dot{x}^{2} + \dot{y}^{2} - x^{2} - y^{2} \right)
[/tex]

harmonic oscillator, and

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

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

[tex]
V(x,y, \dot{x}, \dot{y}) = \frac{1}{2} \left( x \dot{y} - y \dot{x} + x^{2} + y^{2} \right)
[/tex]

To derive the dynamical equations from this potential, one uses

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

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


sam
 
  • #30
samalkhaiat said:
Didn't you learn from elementary mechanics that

[tex] L = T - V[/tex]

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.
 
  • #31
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.

[tex]
A_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle
[/tex]

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:

[tex]
A_{\textrm{classical}}B_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle \langle \Psi|B|\Psi\rangle
[/tex]

And the classical values commute normally. So indeed, how do the Grassmann numbers enter this business?
 
Last edited:
  • #32
jostpuur said:
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! .
 
  • #33
jostpuur said:
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.

[tex]
A_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle
[/tex]

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:

[tex]
A_{\textrm{classical}}B_{\textrm{classical}} = \langle \Psi|A|\Psi\rangle \langle \Psi|B|\Psi\rangle
[/tex]

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

There is no classical physical analogue for Grassmann numbers
 
  • #34
Rainbow Child said:
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 [itex]\hbar\to 0[/itex] out of the operators, but I'm not so convinced.
 
  • #35
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.

samalkhaiat said:
reilly said:
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
 
Last edited:

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