Poisson Brackets, Commutators, and Plane Waves

[topsquark]

Okay, I'm a geek with a lot of time on my hands, so I'm going through all the problems in Sakuri.

The problem: Calculate $$[x^2,p^2]$$. Simple enough. There are basically two fundamental attacks to do this.
1. Direct computation. I get that
$$[x^2,p^2]=2i \hbar (xp+px)$$,
which I got both by slightly fancy means, as well as calculating "xxpp-ppxx" and doing a lot of adding and subtracting to simplify. I'm confident that this answer is correct. (Just to prove everything to a ridiculous level of confidence, I have used a Gaussian distribution to confirm the result by a direct calculation of $$\left \langle [x^2,p^2] \right \rangle$$.)

2. Calculate the Poisson bracket:
$$[x^2,p^2]_P = 4xp$$.
Since we have the correspondence
$$[ \, , \, ]_{QM} \rightarrow i \hbar [ \, , \, ]_P$$
we essentially get the same expression, using the usual "symmetrization" trick that $$2xp \rightarrow xp+px$$.

I am more or less satisfied that both methods give equivalent answers. But, of course, I'm not satisfied with "more or less" and have to go and ruin it. Notice that we don't need to leave the "px" term from the quantum calculation alone. Using the commutator relation: $$[x,p]=xp-px=i \hbar$$ we can get: $$2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2$$. THIS form does not conform to the correspondence between the Poisson bracket and the commutator.

So my first question is: When do we know to leave well enough alone when using the correspondence?

Now, I consider myself to be a realist and (when doing this problem) argued that since the correspondence was clear before I added a purely quantum condition (the [x,p] commutator relation) there is no contradiction here.

However...

I was recently working on a later problem and using the quantum result for the commutator I have what seems to be a problem.

The problem: Calculate the time dependence for $$\left \langle (\Delta x)^2 \right \rangle$$ for a free particle given that $$\left \langle (\Delta x)^2 \right \rangle$$ is known at t = 0. Also given is that <x> = <p> = 0 at t=0.

The problem is a rather straightforward application of time evolution (if rather longish), so the Physics is easy. The problem is that if I use the quantum version of $$[x^2,p^2]=2i \hbar (xp+px)$$ I get a complex time dependence for $$\left \langle (\Delta x)^2 \right \rangle_t$$. To me, this is clearly unphysical: $$\Delta x$$ is a Hermitian operator and the expectation value should therefore be real as well. (I don't believe I made an error in the time evolution of the operator as my results are consistent if I use either $$[x^2,p^2]=2i \hbar (xp+px)$$ or $$2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2$$. When I use $$[x^2,p^2] = 4i \hbar xp$$ I get a similar answer, but an absence of the imaginary linear time term.)

I believe my calculations are done correctly, except for a possible misapplication of the concepts. That is to say that my Math checks out in each of the ways I set the problem up.

So where, conceptually, am I going wrong?

Thanks!
-Dan

Addendum: I almost forgot to mention, since I didn't put in the derivations you might not realize this: The imaginary linear time term is a direct result of the $$2 \hbar^2$$ term in the commutator. (Or equivalently, the result of the distinction between xp and px.)

 

[Hurkyl]

I think showing your work would help. I think I've done the problem right, and I get a real answer for <X²>. (which would then give a real answer for what you're computing)

And, by the way, I would say that

4ihxp + 2h²

corresponds to the classical 4xp: 2h² is "infinitessimal". More precisely, when you divide by ih and take the limit as h->0, that term vanishes.

 

[topsquark]

I think showing your work would help. I think I've done the problem right, and I get a real answer for <X²>. (which would then give a real answer for what you're computing)

And, by the way, I would say that

4ihxp + 2h²

corresponds to the classical 4xp: 2h² is "infinitessimal". More precisely, when you divide by ih and take the limit as h->0, that term vanishes.

(Sorry for the delay, I haven't been feeling well for the last few days.)

Here it is:

Using the H-picture, obtain $$\left \langle (\Delta x)^2 \right \rangle$$ as a function of time. (Recall that in the original problem: <x>=<p>=0 at t=0.)

Notation: I will use $$\left \langle (\Delta x)^2 \right \rangle$$ for the t=0 version and $$\left \langle (\Delta x)^2 \right \rangle (t)$$ for the time dependent.

The time evolution of $$(\Delta x)^2$$ in the H-picture is given as:
$$e^{iHt/ \hbar}(\Delta x)^2e^{-iHt/ \hbar}$$
We may calculate this using the BHC expansion:
$$e^{iG \lambda}Ae^{-iG \lambda}=A+\frac{1}{1!} (i \lambda)[H,A]+\frac{1}{2!}(i \lambda)^2[H,[H,A]] + ...$$

So I begin by calculating commutators.
$$[H,(\Delta x)^2]=[H,(x-<x>)^2]=[H,x^2]$$ since <x>=0 at t=0.
$$[H,x^2]=\left [ \frac{p^2}{2m}, x^2 \right ]=\frac{1}{2m} [p^2,x^2]=\frac{-2i \hbar xp}{m} - \frac{\hbar^2}{m}$$

$$[H,[H,(\Delta x)^2]]=[H,\frac{-2i \hbar xp}{m} - \frac{\hbar^2}{m}]=[H,\frac{-2i \hbar xp}{m}]=\frac{-2i \hbar xp}{2m^2}[p^2,xp]=\frac{-2 \hbar^2 p^2}{m^2}$$

$$[H,[H,[H,(\Delta x)^2]]]=\frac{-2 \hbar^2}{2m^3}[p^2,p^2]=0$$

So:
$$e^{iHt/ \hbar}(\Delta x)^2e^{-iHt/ \hbar}=(\Delta x)^2+\left ( \frac{it}{\hbar} \right ) \left ( \frac{-2i \hbar xp}{m} - \frac{\hbar^2}{m} \right ) +\frac{1}{2} \left ( \frac{it}{\hbar} \right )^2 \left ( \frac{-2 \hbar^2 p^2}{m^2} \right )$$

Thus:
$$\left \langle (\Delta x)^2 \right \rangle (t)=\left \langle (\Delta x)^2 \right \rangle +\frac{2t}{m} \left \langle xp \right \rangle + \frac{it \hbar}{m} + \frac{t^2}{m^2} \left \langle p^2 \right \rangle$$

Since we are dealing with a free particle it is easy to show that $$\left \langle xp \right \rangle$$ is zero:

$$\psi (x) = Ae^{-i(xp/ \hbar - \omega t)}$$ so

$$\left \langle xp \right \rangle = \int_{-\tau /2}^{\tau /2} \, dx \psi^* \hat{x} \hat{p} \psi$$, using a box normalization.

$$=\int_{-\tau /2}^{\tau /2} \, dx Ae^{i(xp/ \hbar - \omega t)} x \frac{\hbar}{i} \frac{d}{dx} Ae^{-i(xp/ \hbar - \omega t)}$$

$$=-A^2 \frac{\hbar}{i} \frac{i}{\hbar} p \int_{-\tau /2}^{\tau /2} \, dx x = 0$$.

So:
$$\left \langle (\Delta x)^2 \right \rangle (t)= \left \langle (\Delta x)^2 \right \rangle + \frac{it \hbar}{m} + \frac{t^2}{m^2} \left \langle p^2 \right \rangle$$

which contains the complex linear time term that bothers me.

-Dan

 

[Hurkyl]

Since we are dealing with a free particle it is easy to show that $\left \langle xp \right \rangle$ is zero:

That can't be right. If <xp> = 0, then we have:
<px> = 0
<xp> - <px> = 0
<xp - px> = 0
<ih> = 0
ih = 0

 

[topsquark]

That can't be right. If <xp> = 0, then we have:
<px> = 0
<xp> - <px> = 0
<xp - px> = 0
<ih> = 0
ih = 0

I'm afraid that's not correct. Remember that x and p are operators:

$$\left \langle px \right \rangle = \int_{-\tau /2}^{\tau /2} \, dx Ae^{i(xp/ \hbar - \omega t)} \hat{p} \hat{x} Ae^{-i(xp/ \hbar - \omega t)}$$

$$= A^2 \int_{-\tau /2}^{\tau /2} \, dx e^{i(xp/ \hbar - \omega t)} \frac{\hbar}{i} \frac{d}{dx} \left ( x e^{-i(xp/ \hbar - \omega t)} \right )$$

$$=A^2 \frac{\hbar}{i} \int_{-\tau /2}^{\tau /2} \, dx e^{i(xp/ \hbar - \omega t)} e^{-i(xp/ \hbar - \omega t)} \, + \,A^2 \frac{\hbar}{i} \frac{-ip}{\hbar} \int_{-\tau /2}^{\tau /2} \, dx e^{i(xp/ \hbar - \omega t)} x e^{-i(xp/ \hbar - \omega t)}$$

$$=A^2 \frac{\hbar}{i} \int_{-\tau /2}^{\tau /2} \, dx \, + \, \,A^2 \frac{\hbar}{i} \frac{-ip}{\hbar} \int_{-\tau /2}^{\tau /2} \, dx x$$

$$=A^2 \frac{\hbar}{i} A^{-2} +0$$

So $$\left \langle px \right \rangle = \frac{\hbar}{i}$$.

-Dan

 

[topsquark]

I have since found another example where the expectation value has an imaginary component.

The problem: Consider the "correlation function" $$\left \langle x(t) x(0) \right \rangle$$. Evaluate the correlation function explicitly for the ground state of a 1-D SHO.

Define x(0) = x, p(0) = p.
We know that:
$$x(t)=xcos( \omega t)+\frac{p}{m \omega}sin(\omega t)$$

So:
$$\left \langle x(t) x(0) \right \rangle = \left \langle 0|x(t)x|0 \right \rangle$$

$$= ... = \frac{\hbar}{2m \omega}e^{-i \omega t}$$

Obviously something is wrong with my concept that the expectation value of a Hermitian operator is always real. In both examples (this one and my original post) time evolution comes into play. I still can't figure it, but is there some connection to time evolution and a complex expectation value?

-Dan

 

[Physics Monkey]

topsquark,

Try taking another look at what Hurkyl said. In detail, $$\langle x p \rangle^* = \langle (x p)^\dag \rangle = \langle p^\dag x^\dag \rangle = \langle p x \rangle$$, and thus if $$\langle x p \rangle = 0$$ then $$\langle p x \rangle = 0$$ which clearly violates the commutation relation.

You may then ask where the mistake is in your derivation. The answer is that you have neglected the fact that the wavefunction is discontinuous at $$x = \pm r/2$$ and the derivative will produce delta functions. A careful treatment (lots of delta functions and step functions running around) will give a nonzero result. Alternatively, if you consider periodic boundary conditions then you have to be more careful with the position operator since it is no longer single valued.

Hope his helps.

 

[topsquark]

topsquark,

Try taking another look at what Hurkyl said. In detail, $$\langle x p \rangle^* = \langle (x p)^\dag \rangle = \langle p^\dag x^\dag \rangle = \langle p x \rangle$$, and thus if $$\langle x p \rangle = 0$$ then $$\langle p x \rangle = 0$$ which clearly violates the commutation relation.

You may then ask where the mistake is in your derivation. The answer is that you have neglected the fact that the wavefunction is discontinuous at $$x = \pm r/2$$ and the derivative will produce delta functions. A careful treatment (lots of delta functions and step functions running around) will give a nonzero result. Alternatively, if you consider periodic boundary conditions then you have to be more careful with the position operator since it is no longer single valued.

Hope his helps.

That IS interesting. (I hate boundary conditions! ) Okay, so we all (now) agree that $$\langle xp \rangle$$ can't be zero. If I am then correct, we can state that the imaginary part of $$\langle xp \rangle$$ is $$i \hbar /2$$ from the commutator relation. Which if you put it into my original derivation, doubles the imaginary linear time term. And, of course we have a "new" imaginary expectation value: $$\langle xp \rangle = a+i \hbar /2$$.

On this thread there have now been listed 3 expectation values of Hermitian operators that have complex values. (And, after thinking about it all day I know of another situation where this occurs.) I can trace the math and see where the imaginary numbers are coming from now: it would appear that the presence of the imaginary numbers can essentially be traced back to the [x,p] commutator. So I guess I have to conclude that some Hermitian operators produce complex expectation values on (at least some) states, even though they are observables.

That seems to me to be rather bizzare. I suppose the difficulty is that most of my coursework focussed on expectation values on eigenstates of a Hermitian operator, which definitely would be a real number, so I'm a little out of my comfort zone here.

Thank you all for your help.

-Dan

 

[Physics Monkey]

Hi topsquark,

Don't give up your faith in the reality of hermitian operators so easily!

In your derivation of the of the time development of $$x^2$$ you missed a sign. The third term should read $$- \frac{i \hbar t}{m}$$ thus canceling the imaginary part of the second term $$\frac{2 t \langle x p \rangle}{m}$$ as it must. You can write the corrected result in an explicitly real form as $$\langle x^2 \rangle + \frac{t}{m} \langle (xp + px) \rangle + \frac{t^2}{m^2} \langle p^2 \rangle$$, a result which is most easily obtained by simply squaring the Heisenberg position operator $$x(t) = x + \frac{p t}{m}$$.

The operator $$xp$$ is not hermitian! Neither is $$x(t) x(0)$$!

 

[topsquark]

Hi topsquark,

Don't give up your faith in the reality of hermitian operators so easily!

In your derivation of the of the time development of $$x^2$$ you missed a sign. The third term should read $$- \frac{i \hbar t}{m}$$ thus canceling the imaginary part of the second term $$\frac{2 t \langle x p \rangle}{m}$$ as it must. You can write the corrected result in an explicitly real form as $$\langle x^2 \rangle + \frac{t}{m} \langle (xp + px) \rangle + \frac{t^2}{m^2} \langle p^2 \rangle$$, a result which is most easily obtained by simply squaring the Heisenberg position operator $$x(t) = x + \frac{p t}{m}$$.

The operator $$xp$$ is not hermitian! Neither is $$x(t) x(0)$$!

Bugger. NOW I know what I was doing wrong. The product of two UNITARY operators is unitary, not the same for Hermitian.

And as for x(t) I was (for some crazy reason) thinking that form was only good for the SHO. Time to reread the section on Ehrenfest's Theorem!

Thanks again!

-Dan