How Does Time Evolution Affect Quantum Measurement Probabilities?

[mcheung4]

Homework Statement

A quantum system has Hamiltonian H with normalised eigenstates ψ_n and corresponding energies E_n (n = 1,2,3...). A linear operator Q is defined by its action on these states:

Qψ₁ = ψ₂
Qψ₂ = ψ₁
Qψ_n = 0, n>2

Show that Q has eigenvalues 1 and -1 and find the corresponding normalised eigenstates ζ₁ and ζ₂, in terms of energy eigenstates. Calculate <H> in each of the states ζ₁ and ζ₂.

A measurement of Q is made at time=0, and the result 1 is obtained. The system is then left undisturbed for a time t, at which instant another measurement of Q is made. What is the probability that the result will again be 1? Show that the probability is 0 if the measurement is made after a time T = \piħ/(E₂ - E₁), assuming E₂ - E₁> 0.

Homework Equations

The Attempt at a Solution

I found
ζ₁ = (ψ₁ + ψ₂)/√2
ζ₂ = (ψ₁ - ψ₂)/√2
and <H> = (E₁ + E₂)/2 for both.

I have trouble doing the second part.

Doesnt the system collapse into ζ₁ given we know this is the state at time =0?
so probability will be 1?

 

[ShayanJ]

The wave function collapses to \zeta_1 after measurement but after that, it will evolve by Schrödinger equation and so it will change gradually. You should apply the time evolution operator to the state \zeta_1 and then calculate the inner product of the evolved state with \zeta_1.

 

[mcheung4]

The wave function collapses to \zeta_1 after measurement but after that, it will evolve by Schrödinger equation and so it will change gradually. You should apply the time evolution operator to the state \zeta_1 and then calculate the inner product of the evolved state with \zeta_1.

so ψ(x,0) = (ψ1 + ψ2)/√2,
and U(t,0) = exp(-iHt/ħ),
then ψ(x,t) = U(t,0)ψ(x,0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2

inner product :
∫ [exp(iHt/ħ)((ψ1)* + (ψ2)*)/√2][(ψ1 + ψ2)/√2] dx
= exp(iHt/ħ)

is this correct? but i don't know what H is here, is it (E1+E2)/2?

 

[ShayanJ]

so ψ(x,0) = (ψ1 + ψ2)/√2,
and U(t,0) = exp(-iHt/ħ),
then ψ(x,t) = U(t,0)ψ(x,0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2

inner product :
∫ [exp(iHt/ħ)((ψ1)* + (ψ2)*)/√2][(ψ1 + ψ2)/√2] dx
= exp(iHt/ħ)

is this correct? but i don't know what H is here, is it (E1+E2)/2?

The point you're missing, is how to act on \psis by the evolution operator.
<br /> e^{\frac{iHt}{\hbar}}\psi_n=\sum_{m=0}^{\infty} (\frac{iH_nt}{\hbar})^m \frac{ \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{it}{\hbar})^m \frac{H^m \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{it}{\hbar})^m \frac{E_n^m \psi_n}{m!}=\sum_{m=0}^{\infty} (\frac{iE_nt}{\hbar})^m \frac{ \psi_n}{m!}=e^{\frac{iE_n t}{\hbar}}\psi_n<br />

So you don't need to know H!

 

[mcheung4]

so ψ(0) = (ψ1 + ψ2)/√2, ψ(t) = U(t,0)ψ(0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 = [exp(-iE1t/ħ)ψ1 + exp(-iE2t/ħ)ψ]/√2. then find the probability amplitude of (ψ1 + ψ2)/√2 in this time-varying state by taking the inner product ((ψ1 + ψ2)/√2 , ψ(t)), squared it to find the probability = [1 + cos((E2-E1)t/h)]/2, then sub in T = pi*h/(E2-E1) indeed got probability = 0. but why after T it'll stay 0? doesn't it cycle again by the consine term?

 

[Ray Vickson]

so ψ(0) = (ψ1 + ψ2)/√2, ψ(t) = U(t,0)ψ(0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 = [exp(-iE1t/ħ)ψ1 + exp(-iE2t/ħ)ψ]/√2. then find the probability amplitude of (ψ1 + ψ2)/√2 in this time-varying state by taking the inner product ((ψ1 + ψ2)/√2 , ψ(t)), squared it to find the probability = [1 + cos((E2-E1)t/h)]/2, then sub in T = pi*h/(E2-E1) indeed got probability = 0. but why after T it'll stay 0? doesn't it cycle again by the consine term?

Why do you say it remains zero? Just because it becomes 0 momentarily does not mean it stays at 0.

 

[nrqed]

so ψ(0) = (ψ1 + ψ2)/√2, ψ(t) = U(t,0)ψ(0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 = [exp(-iE1t/ħ)ψ1 + exp(-iE2t/ħ)ψ]/√2. then find the probability amplitude of (ψ1 + ψ2)/√2 in this time-varying state by taking the inner product ((ψ1 + ψ2)/√2 , ψ(t)), squared it to find the probability = [1 + cos((E2-E1)t/h)]/2, then sub in T = pi*h/(E2-E1) indeed got probability = 0. but why after T it'll stay 0? doesn't it cycle again by the consine term?

I think you simply misread the question but I can see how the phrasing may lead to confusion. What they meant is that at the time T = pi*h/(E2-E1) after the initial measurement was made the probability is zero, but they do not mean to say that it will remain zero for all times after that instant. They really just meant at that specific instant, not at all later times. So you are right, it will cycle again.

 

[mcheung4]

ok thanks guys!