Using the Virial Theorem to Calculate Expectation Values in Quantum Mechanics

[Lindsayyyy]

Hi everybody

Homework Statement

I showed the following:

[tex] \frac{1}{\hbar i} [H,xp]=x \frac {dV}{dx} - \frac {p^2}{m}[/tex]

Now I want to use this to show that:

[tex] \langle \frac {p^2}{2m} \rangle = \frac{1}{2} \langle x \frac {dV}{dx}\rangle[/tex]

whereas |E> is an eigenstate to H with the eigenvalue E. H x and p are operators. Furthermore in my 2nd equation the <> are the expectation value.

Homework Equations

The Attempt at a Solution

Can anyone help me with this? I'm learning for my final exams and I found this task in an old exam. I tried to solve <E|H|E> but that didn't lead me anywhere near it.

Thanks for your help

 

[mfb]

Can you show <[H,xp]> = 0?
I don't know if that helps:
<[H,xp]> = <x[H,p] + [H,x]p>

 

[Lindsayyyy]

[H,xp]= 0 ??

I had to show the first equation which I solved. [H,xp] wasnt 0 there. Or am I misunderstanding something?

edit: oops, is just saw that you mean the expectation value. How to I calculate the expectation value of a commutator? Do I have to solve <E|[H,xp]|E> ?

 

[mfb]

[H,xp] does not have to be 0 everywhere, but its expectation value should vanish if the virial theorem holds.
This directly follows from the virial theorem and your equation, by the way.

 

[andrien]

can you show that expectation of commutator will be
d/dt<x.p>
now <x.p> for stationary state is independent of time.

 

[Lindsayyyy]

I tired what mfb said to show that the expectation value is zero

[tex] \langle E \mid [H,xp] \mid E \rangle=\langle E \mid Hxp\mid E \rangle - \langle E \mid xpH \mid E \rangle = E(\langle E \mid xp \mid E \rangle -\langle E \mid xp \mid E \rangle) [/tex]

Is that correct? If so I showed that the expectation value of the commuator is zero, but it still don't understand how this helps me to get to the 2nd equation in my first post.

Thanks for your help

 

[mfb]

@Lindsayyyy: Take the expectation value of both sides of the first equation, and use linearity of the expectation value? <a+b>=<a>+<b>

 

[Lindsayyyy]

I'm too stupid to solve it,sorry.

I get zero for the left side as I told above but for the right side I got:

[tex] 0=\langle E \mid x \frac {dV}{dx} \mid E \rangle -\langle E \mid \frac {p^2}{m} \mid E \rangle[/tex]

I mean I can just multiply 1/2 on both sides to get to the solution now. But I don't think that's the point here

thanks for your help

 

[mfb]

There is no <E|.|E> in the expectation value, unless you know that you consider the state E only (where did you say that?). And a multiplication with 1/2 gives you the solution, indeed.

 

[Lindsayyyy]

I thought when I want to calculate an expectation value in quantum mechanics it has always the form of:

<n|H|n>

the task says that E is an eigenstate to H with eigenvalue E, that's why I did <E|[H,xp]|E>.
Or how do I calculate an expectation value of [H,xp] to show it is zero?

thanks for your help

 

[mfb]

the task says that E is an eigenstate to H with eigenvalue E

But that is just one eigenstate. What happens if the particle is in a different state?
Do you have the full problem statement somewhere?

 

[Lindsayyyy]

I try to translate it, as I'm not a native speaker and the task is not in english:

The Hamiltonian operator
[tex] \hat{H}= \frac {\hat{p}^2}{2m}+V(\hat{x}) [/tex]

describes a particle in one dimension.

(i)Show the following:

[tex]\frac{1}{\hbar i} [\hat {H},\hat {x}\hat{p}]=\hat {x} \frac {dV}{dx} - \frac {\hat {p}^2}{m}[/tex]

(ii) |E> is an eigenstate of H with the eigenvalue E.Deduce from the expression from task (i) that

[tex] \langle \frac {\hat{p}^2}{2m}\rangle_E =\frac {1}{2} \langle \hat{x} \frac{dV}{dx} \rangle_E[/tex]

whereas [tex] \langle...\rangle_E[/tex] describes the expectation value to the quantum state |E>

I hope that helps.

Thanks for the help and sorry for the sloppy english:shy:

 

[mfb]

Ok, just the expectation value for E. In that case, it is fine.
<.>_E is just <E|.|E>.

 

[Lindsayyyy]

So my solution was right?

But I still don't understand the factor 1/2. Why is it even there? I mean it would be the same without it, wouldn't it?

 

[mfb]

Yes.
You can multiply both sides by 2, if you like, and the equation is still true. ##<\frac{p^2}{2m}>## is the expectation value of the kinetic energy, so usually the factor of 1/2 is included.

 

[andrien]

can you show that expectation of commutator will be
d/dt<x.p>
now <x.p> for stationary state is independent of time.

let us take the right way of doing it.
for any quantum mechanical operator following relation holds
d/dt<O>=(i/h^-)(<[H,O]>)+<∂O/∂t>
now with O=x.p,which does not contain time explicitly last one is zero.
so we have
d/dt<O>=(i/h^-)(<[H,O]>)
now<x(t).p(t)>=<n|e^-iEnt x.p e^iEnt|n>=<x(0).p(0)> so it's independent of time.And the required form follows.

 

[Lindsayyyy]

Ok thanks for the help. I have one more task for this problem.


V is now [tex] V(\lambda x)=\lambda^n V(x)[/tex]

I shall show the following:

[tex] \langle T \rangle_E= \frac {n}{2} \langle V \rangle_E[/tex]

I tried to derive the V for lambda, but that didn't work out. I think I need (ii) from the post above, but I don't get rid of the x operator.

Thanks for your help

 

[andrien]

you can see page 154 here
http://books.google.co.in/books?id=173vjxhvrZoC&pg=PA154&dq=virial+theorem&hl=en&sa=X&ei=6WQLUbvbA4vJrAeM2IDwDA&ved=0CEkQuwUwBA#v=onepage&q=virial%20theorem&f=false

 

[andrien]

uh,ok you can see page 4 here
https://docs.google.com/viewer?a=v&q=cache:OvCO8GhMIZkJ:www.damtp.cam.ac.uk/user/rrh/notes/aqm_notes_180112.pdf+&hl=en&gl=in&pid=bl&srcid=ADGEEShdAgtSTmNEh3X6fkcfef_CDk9cFKvdrPeAkauFTJRd8IUrBgZDymoGbj89F72kSzA8r7fi8K275ACqzVuEYPH-owL-DY2Zvb2pMYSmJcUW0SDBTsVPj9vfvob8BLBLmnJQByF1&sig=AHIEtbRigSSpyQh9UlkY0dWtTJmi1OPr3Q

 

[Lindsayyyy]

If anyone in the future will have the same problem and might find this thread, I solved the problem in a much more (in my opinion) easier way than in the links state above with the euler theorem

we have the potential [tex] V(\lambda x)=\lambda^n V(x)[/tex]

derive on both sides wheres I substutite lambda*x=u

[tex] \frac {dV}{du} \frac {du}{d\lambda}= n \lambda^{(n-1)}V(x)[/tex]

left side equals
[tex] \frac {dV}{du} x[/tex]

set lambda to 1 and you get

[tex] \frac {dV}{dx} x = n V(x)[/tex]