Why Does the Integral of Commutator [xp + px] Between Degenerate States Vanish?

[klw1026]

Schiff QM chapter 3 prob 9
Let u1 (r) and u2 (r) be two eigenfunctions of the same hamiltonian that correspond to the same energy eigenvalue; they may be the same function, or they may be degenerate. Show that

int[u∗1 (r) [xp + px] u2 (r)]dr = 0​

where the momentum operator p = (−i/hbar)(∂ /∂ x) operates on everything to its right.

Another eager mind and myself have been working on this problem for two weeks. We cannot seem to get the right things to line up. We have tried integration by parts, we have used some assumptions on the structure of the eigenfunctions, and we have even used the following identity in order to try and incorporate the hamiltonian into the integral

d/dt(<x^2>)=(1/m)(<xp>+<px>) where <> are the expectation value signs;​

but this fails since the eigenfunctions are of the stationary equation and therefore
independent of time. If anyone could help please let us know.

 

[klw1026]

test
test
test

 

[variation]

Hello,

In yuor question, you do not metion that the u_i(r) are three-dimensional or one-dimensional. I make it simply and take it as a one-dimensional problem. If it is three dimensional one, the following still works.

u_1 and u_2 are corresponding to the same eigenvalue.
The aim of the problem is
\int u_1(x)\left(x\frac{\hbar}{i}\frac{\partial}{\partial x}+\frac{\hbar}{i}\frac{\partial}{\partial x}x\right)u_2(x)dx=0.
Therefore this should be satisfied, too.
\int u_2(x)\left(x\frac{\hbar}{i}\frac{\partial}{\partial x}+\frac{\hbar}{i}\frac{\partial}{\partial x}x\right)u_1(x)dx=0.
One can sum of the left hand side of the two equations, integrate by part, and obtain that the sum vanishes.
I drop the detail calculation above.
But logivally speaking, a prerequisite should be satisfied in my method:
" FOR ANY TWO "u_1 and u_2 of the same eigenenergy,...



Hope these helpful

 

[klw1026]

Dear Variation,
I do not follow what you are trying to do. Are you suggesting to add the two equations together? Keep in mind the eigenfunctions to the left of the two operators are complex conjugates. Can you please expand your details some more, if at all possible. Also, how does this method utilize the fact that the eigenfunctions belong to the same energy eigenvalue? Thank you for the time you spent on the problem.

 

[variation]

Hello,

Yes, ok! Sorry, I take the box as nothng.
So, it is complex conjugate.
Thank you

 

[klw1026]

Yes, everything you wrote in your first reply is correct, except that U1(x) is the complex conjugate. Also, how are you getting those beautiful mathematical fonts into the blog? I wrote the question in LaTeX but it did not transfer into the blog correctly.

 

[klw1026]

Of course I mean the first equation is where U1(x) is the conjugate, in the second equation it would be U2(x). Sorry!

 

[neelakash]

Is this problem at all done?

I am trying to do this problem and another of like category and posted here in this forum.

But I am not finfing this to be helpful.

 

[neelakash]

Integration by parts will not help as one cannot discard some terms by demandiong that they vanish at infinity.

I+I=2I= ∫ u1* (xp+px) u2 dx + ∫ u2* (xp+px) u1 dx

= ∫ u1* xp u2 dx + ∫ u2* xp u1 dx + ∫ u1* px u2 dx + ∫ u2* px u1 dx

= (u1, xp u2)+ (u2, xp u1)+ (u1, px u2)+ (u2, px u1)

= (px u1, u2)+ (px u2, u1)+ (u1, px u2)+ (u2, px u1)

Now this should be zero...But how to prove that?