Thursday, June 6, 2019

hilbert space - Use of operators in a time-dependent Hamiltonian quantum system


I am given the following Hamiltonian, H=H1=p22m+12mω21x2

for t<0 and H=H2=p22m+12mω22x2
for t0. For some time t1(<0), we know the eigenstates of the system as {|n1} which are eigenstates of H1 and for some time t2(>0), we know the eigenstates of the system as {|n2} which are eigenstates of H2.


Now, my question is, at time t1, we have a whole list of operators which can act on the state like a1 and a1 i.e the annihilation and creation operators and similarly at time t2 as a2 and a2. Now, if I act a1 on |02, or perform similar operations, what do I get? Will the eigenvectors of a1 also be the eigenvectors of a2?



Answer



a1 is not hermitian so be careful about its eigenvectors. Most importantly, the definition of the creation and destruction operators involves the frequency ω so the creation operator on both sides of t=0 will not be the same so the harmonic oscillator eigenstates, i.e. the number states, will not be the same. This is evident because the length scale for the number states depends also on ω.


In particular, the change of scale is equivalent to a squeezing transformation so the vacuum state for t2 will not be the vacuum state for t1. You will need to expand |02 in terms of t1-states in order compute the action of a1, i.e. given λ1=mω1 ψn(x)=12nn!(λ21π)1/4eλ21x2/2Hn(λ1x)

and ϕn(x)=12nn!(λ22π)1/4eλ22x2/2Hn(λ2x)
the simplest way to proceed is to write |n2=n1|n1n1|n2
with n1|n2=ϕn2(λ2x)ϕn1(λ1x)
Thus, for instance, 01|02=2λ1λ2λ21+λ22,21|02=λ1λ2(λ21λ22)(λ21+λ22)3/2


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