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Commutation and eigenstates

by hokhani
Tags: commutation, eigenstates
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hokhani
#1
Mar3-12, 01:36 PM
P: 269
We now that if [A,B]=0, they have the same eigenstates. But consider a harmonic oscillator with the spring constant k1. If we change k1 to k2, then [H1,H2]=0 and the above expression implies that the eigenstates should not change while they really change!
Could you please tell me if i am wrong?
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kof9595995
#2
Mar3-12, 02:28 PM
P: 679
I don't think [H1,H2] is 0, [H1,H2]=[p^2+k_1*x^2, p^2+k_2*x^2]=k_1[x^2,p^2]+k_2[p^2,x^2]=(k_1-k_2)[x^2,p^2], neither of the two factors is 0.
tom.stoer
#3
Mar4-12, 03:42 AM
Sci Advisor
P: 5,443
1) kof9595995 is right, the two Hamiltonian do not commute b/c H2 = H1 + (k2-k1)x2 and therefore [H1,H2] = [H1,H1 + (k2-k1)x2] = [H1,H1 + (k2-k1)x2] = (k2-k1)/2m [p2,x2]

2) the eigenvectors for two commuting observables are relevant for two observables in one system; but two Hamiltonians describe two different systems, so this is an academic question

M Quack
#4
Mar4-12, 04:17 AM
P: 667
Commutation and eigenstates

1) agree

2) In the sudden approximation you apply a new Hamiltonian to an "old" set of wave functions to find the transition probabilities due to the change. If the "new" Eigenvectors are the same as the old ones, you know exactly where you end up.
tom.stoer
#5
Mar4-12, 04:20 AM
Sci Advisor
P: 5,443
regarding 2) there may indeed be some applications where this makes sense; but in general you are intersted in a system described by a unique H and by a set of observables {H, O1, O2, ...}.


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