Let u1 (x) and u2 (x) be two eigenfunctions of the same hamiltonian that correspond to the same energy eigenvalue;Show that

∫ u1* (xp+px) u2 dx =0

Attempt 1.

Let I=∫ u1* (xp+px) u2 dx =(u1, A u2) where A=(xp+px) is hermitian.

Now as u1 and u2 are arbitrary, we should also have I=∫ u2* (xp+px) u1 dx

Then, 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???

Attempt 2.

If u1 and u2 are eigenfunction of the Hermitian operator A then they are likely to be non-degenerate.And then from the theorem, the eigenfunctions are orthogonal and the integral is zero.

But, H and (xp+px) do not commute!!! So, u1 and u2 are not in general eigen functions of A

Can anyone please suggest any method to do this problem???

Perhaps constructing something which leads to xp-px would be more useful, since that reduces to ih' and then that would turn the integral into ∫ u1* u2, which you know to be zero due to orthogonality of the degenerate eigenfunctions.

If that was the problem,then I would not have asked it here.Note that my problem includes (xp+px) not (xp-px).Also in that case, u1 and u2 are the eigenfunctions corresponding to the same eigenvalue.So,we cannot use the famous theorem so easily.

In our university model questions booklet, an equivalent problem appears like this:

Let u1 (x) and u2 (x) be two eigenfunctions of the same hamiltonian that correspond to the same energy eigenvalue;Show that

∫ u1 (xp-px) u2 dx =0 [u1 and u2 are orthogonal]

Note that here they give u1 but not u1*. So,I am confused which between the two is correct...if at all.may be both problems are correct.But I am not having any clue.

It looks either in this problem (2nd one) nothing is to be done or there is something subtle that I am missing and it makes both the problems meaningful.

In atleast 3-4 different forums,people are avoiding this problem.

I need to know whther at all this problem is worth of attention any further.If these two are correct problems, I can try again.But I do not know if they are.

Surprisingly,our Professor in College also avoided by suggesting the trivial answer.

Can anyone say if these problems are correct?

Hi kolahal_b,

Both problems are incorrect.

In the problem you first gave, the answer is clearly nonzero. This is because when you have two eigenfunctions with the same energy eigenvalue, they are not automatically orthogonal. As you stated, this theorem applies when the eigenvalues are different and the operator is Hermitian. In the case of two eigenfunctions of the same eigenvalue, they MAY BE CHOSEN to be orthogonal, but this is not an automatic consequence of their being two different eigenfunctions. Usually they are chosen to be orthogonal, but as this is a choice and not a theorem, it cannot be used in a proof unless it is explicitly stated as an assumption. The problem does not do that. Because of this, I could take any two eigenfunctions of the same eigenvalue E, |E,1> and |E,2> and form the linear combination a|E,1> + b|E,2> with a very close to 1. This would then certainly not be orthogonal to |E,1>. In fact, the problem does not even state that these are two DIFFERENT eigenfunctions, so it is allowed that they are the same.

In that case, consider the free space Hamiltonian, and the eigenfunction exp(ikx). The operator xp+px gives h/(pi)*2xk -ih/2pi , and exp(ikx) multiplied by its conjugate gives 1, so the integrand becomes h/pi*2xk - ih/2pi, which diverges when integrated between infinite limits. Even with box normalization, the integral is clearly nonzero.

As for the second problem, it has two errors. First, it certainly should be the complex conjugate of the left eigenfunction. Second, since it is known that (as Alphanumeric mentioned) xp-px = ih/2pi, and since this is a non-differential operator which simply multiplies by a constant, the integrand becomes ih/2pi u1(x) u2(x). Even with the conjugate of u1, the value of this integral could be anything, because u1 and u2 are not necessarily orthogonal. If you are supposed to assume that they are orthogonal, then the problem is too easy (except for the missing conjugate) because it reduces down to just knowing that xp-px is a constant. On the other hand, if you are supposed to prove by this calculation that u1 and u2 are orthogonal, than it is impossible, since they may easily be non-orthogonal because they have the same eigenvalue. Obviously you cannot be expected to prove something that isn't even always true!

I think you should spend no more time on these problems. They are not worth your effort.

Hope that helps!

--Stuart Anderson

Thank you very much.It is llike a headache reduced...