How Does the Feynman-Hellmann Theorem Apply to Quantum Systems?

[cepheid]

Homework Statement

Suppose the Hamiltonian H for a particular quantum system is a function of some parameter λ. Let E_n(λ) and ψ_n(λ) be the eigenvalues and eigenfunctions of H(λ). The Feynman-Hellmann Theorem states that

\frac{\partial E_n}{\partial \lambda} = \left \langle \psi_n \left | \frac{\partial H}{\partial \lambda} \right | \psi_n \right \rangle​

assuming that either E_n is non-degnerate or --- if degenerate, that the ψ_n's are the "good" linear combinations of the degenerate eigenfunctions.

a) Prove the Feynman-Hellman Theorem Hint: Use Equation 6.9
b) Apply it to the 1D Harmonic Oscillator (i) using λ=ω (this yields a formula for <V>), (ii) using λ = ħ (this yields <T>), and (iii) using λ = m (this yields a relation between <T> and <V>.

Homework Equations

Equation 6.9 is for the first order correction to the energy, given H' is the perturbation:

E_n^1 = \langle \psi_n^0 | H^\prime | \psi_n^0 \rangle​

The Attempt at a Solution

I said to represent a perturbation as a small change in λ:

H(\lambda + \delta \lambda) = H(\lambda) + \frac{\partial H}{\partial \lambda} \delta \lambda​

to first order.

Similarly

E_n(\lambda + \delta \lambda) = E_n(\lambda) + \frac{\partial E_n}{\partial \lambda} \delta \lambda​

So, since the first order term in the expansion of the Hamiltonian above is just H', and the first order term in the expansion of the energy above is just E¹_n, equation 6.9 becomes:

\frac{\partial E_n}{\partial \lambda} \delta \lambda = \left \langle \psi_n \left | \frac{\partial H}{\partial \lambda} \delta \lambda \right | \psi_n \right \rangle​

Then I just divided both sides by δλ to get the result. What I'm wondering is, I just made this up. Is it a valid proof? Wikipedia has another method that makes use of the definition E = <H> and nothing else:

http://en.wikipedia.org/wiki/Hellmann-Feynman_theorem

However, this method doesn't use the given equation 6.9. That's all I'm asking about for now. I have a question about part b as well, but I'll wait until we get part a out of the way.

 

[kdv]

Your way of doing it sounds completely right to me.

 

[cepheid]

Thanks for weighing in on that kdv. I have a question about part (b) which is as follows:

Apply it (the Feynman-Hellman theorem) to the one-dimensional harmonic oscillator, (i) using λ =ω (this yields a formula for <V>).

So we have:

E_n = \left(n+\frac{1}{2}\right)\hbar \omega

\frac{\partial E_n}{\partial \omega} = \left \langle \psi_n \left | \frac{\partial H}{\partial \omega } \right | \psi_n \right \rangle

\left(n+\frac{1}{2}\right)\hbar = \left \langle \psi_n \left | \frac{\partial }{\partial \omega }\left (-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \right) + \frac{\partial}{\partial \omega}\left(\frac{1}{2}m\omega^2x^2\right) \right | \psi_n \right \rangle

bear with me (continued in next post)

 

[cepheid]

When I was doing this problem, my classmates said that the first term in the central portion of the right hand side (dT/dw) should be zero, because "that part of the Hamiltonian doesn't depend on omega". To me, however, differentiating T with respect to omega is meaningless. T is a differential operator. The only reasonable way for me to interpret this:

\frac{\partial H}{\partial \omega}

is that it is itself an operator of the form AB, where A = d/dw and B = H. Applying this operator to | \psi_n \rangle consists of first having H act on psi, and then having d/dw act on the RESULT of that. Is that true? Does...

\frac{\partial H}{\partial \omega}|\psi_n \rangle = \frac{\partial}{\partial \omega}(H |\psi_n \rangle)

If so, that hugely complicates the calculation. You can't just differentiate V wrt omega and you don't get an expression for <V> as Griffiths claimed you would. I guess I just don't understand the math well enough to deal with this problem. If you really are allowed to differentiate H first, before it acts on anything, what the hell does that MEAN?

 

[kdv]

When I was doing this problem, my classmates said that the first term in the central portion of the right hand side (dT/dw) should be zero, because "that part of the Hamiltonian doesn't depend on omega". To me, however, differentiating T with respect to omega is meaningless. T is a differential operator. The only reasonable way for me to interpret this:

\frac{\partial H}{\partial \omega}

is that it is itself an operator of the form AB, where A = d/dw and B = H. Applying this operator to | \psi_n \rangle consists of first having H act on psi, and then having d/dw act on the RESULT of that. Is that true? Does...

\frac{\partial H}{\partial \omega}|\psi_n \rangle = \frac{\partial}{\partial \omega}(H |\psi_n \rangle)

If so, that hugely complicates the calculation. You can't just differentiate V wrt omega and you don't get an expression for <V> as Griffiths claimed you would. I guess I just don't understand the math well enough to deal with this problem. If you really are allowed to differentiate H first, before it acts on anything, what the hell does that MEAN?

I think that one way to think about it is to take the derivative with respect to the parameter before even quantizing. That is, you take the derivative on the classical hamiltonian. Then you quantize. So it gives that the kinetic operator does not contribute.

 

[cepheid]

I think that one way to think about it is to take the derivative with respect to the parameter before even quantizing. That is, you take the derivative on the classical hamiltonian. Then you quantize. So it gives that the kinetic operator does not contribute.

Hey...now THAT's an interesting possibility I hadn't considered. Does anyone know what the rules are regarding when the canonical substitution is supposed to be made?

Thanks again for weighing in kdv.

 

[cepheid]

Nobody?

 

[genneth]

I think you're overthinking it. There is no reason why you can't take the derivative of an operator, or the derivative of an expression involving an operator.