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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 [tex] [x^2,p^2] [/tex]. Simple enough. There are basically two fundamental attacks to do this.
1. Direct computation. I get that
[tex] [x^2,p^2]=2i \hbar (xp+px) [/tex],
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 [tex] \left \langle [x^2,p^2] \right \rangle [/tex].)
2. Calculate the Poisson bracket:
[tex] [x^2,p^2]_P = 4xp [/tex].
Since we have the correspondence
[tex] [ \, , \, ]_{QM} \rightarrow i \hbar [ \, , \, ]_P [/tex]
we essentially get the same expression, using the usual "symmetrization" trick that [tex]2xp \rightarrow xp+px [/tex].
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: [tex][x,p]=xp-px=i \hbar [/tex] we can get: [tex]2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2 [/tex]. 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 [tex] \left \langle (\Delta x)^2 \right \rangle [/tex] for a free particle given that [tex] \left \langle (\Delta x)^2 \right \rangle [/tex] 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 [tex] [x^2,p^2]=2i \hbar (xp+px) [/tex] I get a complex time dependence for [tex] \left \langle (\Delta x)^2 \right \rangle_t [/tex]. To me, this is clearly unphysical: [tex]\Delta x [/tex] 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 [tex] [x^2,p^2]=2i \hbar (xp+px) [/tex] or [tex]2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2 [/tex]. When I use [tex] [x^2,p^2] = 4i \hbar xp [/tex] 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 [tex]2 \hbar^2 [/tex] term in the commutator. (Or equivalently, the result of the distinction between xp and px.)
The problem: Calculate [tex] [x^2,p^2] [/tex]. Simple enough. There are basically two fundamental attacks to do this.
1. Direct computation. I get that
[tex] [x^2,p^2]=2i \hbar (xp+px) [/tex],
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 [tex] \left \langle [x^2,p^2] \right \rangle [/tex].)
2. Calculate the Poisson bracket:
[tex] [x^2,p^2]_P = 4xp [/tex].
Since we have the correspondence
[tex] [ \, , \, ]_{QM} \rightarrow i \hbar [ \, , \, ]_P [/tex]
we essentially get the same expression, using the usual "symmetrization" trick that [tex]2xp \rightarrow xp+px [/tex].
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: [tex][x,p]=xp-px=i \hbar [/tex] we can get: [tex]2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2 [/tex]. 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 [tex] \left \langle (\Delta x)^2 \right \rangle [/tex] for a free particle given that [tex] \left \langle (\Delta x)^2 \right \rangle [/tex] 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 [tex] [x^2,p^2]=2i \hbar (xp+px) [/tex] I get a complex time dependence for [tex] \left \langle (\Delta x)^2 \right \rangle_t [/tex]. To me, this is clearly unphysical: [tex]\Delta x [/tex] 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 [tex] [x^2,p^2]=2i \hbar (xp+px) [/tex] or [tex]2i \hbar (xp+px) = 4i \hbar xp + 2 \hbar^2 [/tex]. When I use [tex] [x^2,p^2] = 4i \hbar xp [/tex] 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 [tex]2 \hbar^2 [/tex] term in the commutator. (Or equivalently, the result of the distinction between xp and px.)
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