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Solved Problem on Virial Theorem - Quantum Mechanics II – Homework | PHYS 702, Assignments of Quantum Mechanics

Material Type: Assignment; Class: Intro Quantum Mechanics II/Hon; Subject: Physics; University: University of New Hampshire-Main Campus; Term: Fall 2007;

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QUANTUM MECHANICS II
Department of Physics Homework 7
University of New Hampshire Due November 16th, 2007
You may find that these homework problems are getting more involved.
Don’t wait until the last minute trying to solve them!
1. Griffiths 6.12
Use the Virial Theorem to prove that for the Hydrogen atom we have:
1
r=hψnlm|1
r|ψnlmi=1
n2a(1)
Some useful notes:
Remember back from last semester when we derived the Heisenberg uncer-
tainty principle (see page 115 of Griffiths). We found that the rate of change
of the expectation value of some operator Qis given by:
d
dt hQi=i
¯hDhˆ
H, ˆ
QiE+*∂Q
∂t +(2)
If we now choose Q=~r ·~p, (with ~r = (r1, r2, r3) )we get:
d
dt h~r ·~p i=i
¯hDhˆ
H, ~r ·~p iE+*(~r ·~p )
∂t +(3)
But the last term is zero (since neither ~r nor ~p explicitly depend on t(they
are operators!). The first term is also zero for stationary states, i.e. for the
eigenstates of the Hamiltonian. We thus only need to evaluate the middle
term:
[H, ~r ·~p] =
3
X
i=1
[H, ripi] =
3
X
i=1
{ri[H, pi] + [H, ri]pi}= 0 (4)
Working out each term separately:
[H, ri] = "p2
2m, ri#+ [V, ri] = 1
2mhp2, rii=1
2m
3
X
j=1 hp2
j, rii(5)
3
X
j=1 hp2
j, rii=
3
X
j=1
{pj[pj, ri] + [pj, ri]pj}=
3
X
j=1
{pj(i¯ij) + (i¯ij )pj}=2i¯hpi(6)
pf3

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QUANTUM MECHANICS II

Department of Physics Homework 7 University of New Hampshire Due November 16th, 2007

You may find that these homework problems are getting more involved. Don’t wait until the last minute trying to solve them!

  1. Griffiths 6. Use the Virial Theorem to prove that for the Hydrogen atom we have: 〈 (^1)

r

〉 = 〈ψnlm|

r

|ψnlm〉 =

n^2 a

Some useful notes: Remember back from last semester when we derived the Heisenberg uncer- tainty principle (see page 115 of Griffiths). We found that the rate of change of the expectation value of some operator Q is given by: d dt

〈Q〉 =

i ¯h

〈[ H,ˆ Qˆ

]〉

〈 ∂Q ∂t

〉 (2)

If we now choose Q = ~r · ~p, (with ~r = (r 1 , r 2 , r 3 ) )we get: d dt

〈~r · ~p 〉 =

i ¯h

〈[ H, ~ˆ r · ~p

]〉

〈 ∂ (~r · ~p ) ∂t

〉 (3)

But the last term is zero (since neither ~r nor p~ explicitly depend on t (they are operators!). The first term is also zero for stationary states, i.e. for the eigenstates of the Hamiltonian. We thus only need to evaluate the middle term: [H, ~r · ~p] =

∑^3 i=

[H, ripi] =

∑^3 i=

{ri [H, pi] + [H, ri] pi} = 0 (4)

Working out each term separately:

[H, ri] =

[ p^2 2 m

, ri

]

  • [V, ri] =

2 m

[ p^2 , ri

]

2 m

∑^3 j=

[ p^2 j , ri

] (5)

∑^3 j=

[ p^2 j , ri

]

∑^3 j=

{pj [pj , ri] + [pj , ri] pj } =

∑^3 j=

{pj (−i¯hδij ) + (−ihδ¯ij ) pj } = − 2 i¯hpi (6)

[H, pi] =

[ p^2 2 m

, pi

]

  • [V, pi] = i¯h

∂V

∂ri

Putting this all together, we get:

〈[H, ~r · ~p ]〉 =

〈 (^3) ∑ i=

riih¯

∂V

∂ri

2 m

(− 2 i¯hpi) pi

= i¯h

〈 ~r · ∇V −

p^2 m

〉 = 0 (8)

So finally we get: 2 〈T 〉 = 〈~r · ∇V 〉 (9) Which is the Quantum version of the Virial Theorem in 3 dimensions. Now compute ~r · ∇V and use 〈H〉 = 〈T 〉 + 〈V 〉 = E to get your result.

  1. Griffiths 6. Find the lowest order relativistic correction to the energy levels of the one- dimensional harmonic oscillator. (And please, use the ladder operators!).
  2. Griffiths 6. Suppose the Hamiltonian, H, for a particular quantum system, is a function of some parameter λ. The solutions to the Schr¨odinger’s equation then give the eigenvalues E(λ) and the eigenfunctions ψ(λ) of Hψ = Eψ.

(a) Proof the Feynman-Hellmann theorem, which states that:

∂En ∂λ

〈 ψ

∣∣ ∣∣ ∣

∂H

∂λ

∣∣ ∣∣ ∣ ψ

〉 (10)

Hint: We need to use perturbation theory to prove this (note that ψ(λ) so don’t be tempted to just differentiate E = 〈ψ |H| ψ〉.). Let λ 0 be any of the variables in H. If you ”tweak” λ 0 just a little to λ = λ 0 + dλ, work out what the perturbation Hamiltonian H′^ will be. Why is the first order perturbation theory result for E(1)^ exact in this case? (b) Apply the theorem to the one dimensional harmonic operator and com- pare the results with previously obtained answers. Do this for i) λ = ω, ii) λ = ¯h, iii) λ = m. (See your book. Note you can answer this even if you did not get part a).