# Fermion oscillator

1. Jan 8, 2008

### jostpuur

Problem:

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

$$L=\dot{x}y - x\dot{y} - x^2 - y^2 ?$$

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

$$\dot{x}(t) = y(t)$$
$$\dot{y}(t) = -x(t),$$

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

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

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

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

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 SchrÃ¶dinger's equation and Hamiltonian:

The Hamiltonian can be solved to be

$$H=\frac{\partial L}{\partial\dot{x}}\dot{x} + \frac{\partial L}{\partial\dot{y}}\dot{y} - L = x^2 + y^2 = |z|^2$$

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

$$i\partial_t\Psi(t,z) = |z|^2\Psi(t,z),$$

and solutions are

$$\Psi(t,z) = e^{-i|z|^2 t}f(z).$$

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

$$p_x = \frac{\partial L}{\partial\dot{x}} = y \quad\quad\quad p_y = \frac{\partial L}{\partial\dot{y}} = -x$$

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

$$z(t) = \frac{(t_1-t)z' + (t-t_0)z}{t_1-t_0},$$

and compute the action

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

but this cannot be used to define time evolution with

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

as usual, because there is no

$$\propto \quad\frac{1}{t_1-t_0},\quad\quad\textrm{or}\quad \frac{1}{(t_1-t_0)^2}$$

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

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

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

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

So the fermion analogy to the harmonic oscillator is

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

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.

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.

Last edited: Jan 8, 2008
2. Jan 8, 2008

### 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.

3. Jan 8, 2008

### 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

4. Jan 8, 2008

### olgranpappy

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

5. Jan 8, 2008

### 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.

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

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)

6. Jan 9, 2008

### 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

7. Jan 9, 2008

### jostpuur

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?

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

$$L = \dot{x}y - x\dot{y} + x^2 + y^2,$$

the Hamiltonian would be

$$H=-|z|^2,$$

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

$$z(t) = e^{it}z(0)$$

This system doesn't work very intuitively.

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

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

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

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?

Last edited: Jan 9, 2008
8. Jan 9, 2008

### 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.

9. Jan 9, 2008

### olgranpappy

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

10. Jan 9, 2008

### 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

$$H = \boldsymbol{x}\times\boldsymbol{p} + |\boldsymbol{x}|^2?$$

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

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

$$L = \dot{x}y - x\dot{y} - x^2 - y^2$$

$$H = \frac{\partial L}{\partial\dot{x}}\dot{x}\; +\; \frac{\partial L}{\partial\dot{y}}\dot{y} \;-\; L = y\dot{x} \;+\; (-x)\dot{y} \;-\; \big(\dot{x}y \;-\; x\dot{y}\; -\; x^2 \;-\; y^2\big) = x^2 + y^2$$

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

$$H = xp_y - yp_x + x^2 + y^2$$

Last edited: Jan 9, 2008
11. Jan 10, 2008

### Rainbow Child

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 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. Jan 10, 2008

### jostpuur

("Lagrangian of GR is"?)

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

13. Jan 10, 2008

### Rainbow Child

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

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

Yes!

14. Jan 10, 2008

### jostpuur

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

Good piece of information to know.

15. Jan 11, 2008

### samalkhaiat

Last edited: Jan 11, 2008
16. Jan 11, 2008

### samalkhaiat

17. Jan 12, 2008

### jostpuur

Wait a minute. We have

$$\left\{\begin{array}{l} \dot{x} = y\\ \dot{y} = -x\\ \end{array}\right. \quad\implies\quad \left\{\begin{array}{l} \ddot{x} = -x\\ \ddot{y} = -y\\ \end{array}\right.$$

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.

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

https://www.amazon.com/Quantization-Gauge-Systems-Marc-Henneaux/dp/0691037698

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.

Last edited: Jan 12, 2008
18. Jan 12, 2008

### jostpuur

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

$$L=\epsilon(\dot{x}^2 + \dot{y}^2) + \dot{x}y - x\dot{y} - x^2 - y^2$$

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

19. Jan 12, 2008

### samalkhaiat

20. Jan 12, 2008

### Rainbow Child

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.