Solve Quantum Measurement Homework: Wavefunction After q1 Measurement

[hasan_researc]

Homework Statement

For a particle in an isolated system,
the Hamiltonian operator has normalised eigenstates and eigenvalues u_{n}(x) and E_{n}, respectively.
The operator of another variable Q has normalised eigenstates and eigenvalues \phi_{n} and q_{n}, respectively.

The lowest two Q eigenstates happen to be related to those of energy by
\phi_{1}(x) = \frac{\sqrt{2}u_{1}(x) + u_{2}(x)}{\sqrt{3}},
\phi_{1}(x) = \frac{u_{1}(x) - \sqrt{2}u_{2}(x)}{\sqrt{3}}.

A measurement of Q is made at time t = 0 and the result is q_{1}.
What is the wavefunction \psi (x , 0) immediately after the measurement?

Homework Equations

The Attempt at a Solution

I have no idea, really!

 

[vela]

This question pertains to a pretty basic aspect of making a measurement in quantum mechanics. Review what your textbook says about what a measurement does to the wave function.

 

[hasan_researc]

A measurement collapses a wavefunction.

So, \psi(x , 0) = \phi_{1}.

Am I right?

 

[vela]

Exactly. A lot simpler than you thought, huh?

 

[hasan_researc]

But the question doesn't end there. The next part is:

A second measurement is made immediately after the first.

(a) If the second measurement is of Q,
list the possible outcomes of this second measurement and the probability of each.

(b) Alternatively, if the second measurement is of E,
list the possible outcomes of this second measurement and the probability of each.

Shall I try to answer by myself so you can check and comment?

 

[vela]

Yes, give it a shot.

 

[hasan_researc]

(a) After the first measurement:
(the system is isolated, so) assuming the measurement did not affect the system,
possible outcome of the second measurement = q_{1} and
probability of q_{1} = 1.

(b) After the first measurement:
(the system is isolated, so) assuming the measurement did not affect the system,
possible outcomes of the second measurement = E_{1} and E_{2} and
probability of E_{1} = 2/3 and E_{2} = 1/3.

What do you think?

I find my assumptions dubious? Thoughts?

 

[vela]

Your answers are correct.

What do you consider dubious?

 

[hasan_researc]

The system is isolated, so only a measurement can change the state of the system.
In other words, only a measurement can change the eigenvalue and the eigenstate of the system.

Therefore,
(a) the state of the system has been determined by the first measurement.
The second measurement disturbs the system, so should it not change the state of the system?
In other words, shouldn't the eigenvalues and eigenstates of the system change?

(b) the state of the system has been determined by the first measurement.
The second measurement disturbs the system, and we are assuming that the measurement does not affect the system. I don't understand if we need this assumption to get to the answer.

 

[vela]

Performing a measurement can change the state of the system. For example, in this problem, you actually don't know what the state of the system was initially. All you know is if you expanded the wave function of the state in eigenfunctions of Q, the coefficient of ϕ₁ was not zero because q₁ was a possible outcome. When the first measurement yielded the result q₁, you then knew that the wave function of the system collapsed to ϕ₁. The next measurement may similarly cause the state to change, and what it changes to depends on the result of the measurement.

 

[hasan_researc]

I have to confess that I have not fully understand what you wrote, but putting that aside, let's worry about the final part of the question!

In a different experiment,
a first measurement of Q at t = 0 again results in q₁,
but the second measurement is delayed by a time t.

(a) Give an expression for the wavefunction ψ(x , t) just before the second measurement.

(b) If the delayed second measurement is of E,
list the possible outcomes of the second measurement and the probability of each.

(c) Alternatively,
if the delayed second measurement is of Q,
again list the possible outcomes of this second measurement and give expressions for the probability of each.

Shall I first answer by myself so you can check and comment?

 

[vela]

Yes.

 

[hasan_researc]

<br /> \psi(x, t) = \frac{ \sqrt{2}u_{1}(x)e^{-iE_{1}t/hcross} + u_{2}(x)e^{-iE_{2}t/hcross} }{ \sqrt{3} }<br />

What do you think?

 

[vela]

That's correct for part (a).

 

[hasan_researc]

I don't know how to solve for (b). Any ideas?

 

[vela]

The same way you did before by looking at the coefficients of the energy eigenstates.

 

[hasan_researc]

OK, so

(b) possible outcomes of the second measurement = E_{1} and E_{2}
and probability of E_{1} = 2/3 and E_{2} = 1/3.

What do you think?

 

[Arya_svit-kon]

Hey :)
hope you don't mind me joining the post! Only I'm doing the same problems. Also apologies before hand if the text comes through wrong on posts, it's the first time I've used a forum before!
Sorry to go back to an earlier question, but when finding the probabilities of E1 and E2, how did you do it? I know you take "the modulus of an squared", and this will sound ridiculously lame (I am rubbish with quantum mechanics) but how do you get 2/3 and 1/3? because to take the modulus of E1 would be that of (sqrt(2)+1)all squared over 3 ? surely? although to be honest I don't really understand what "an" stands for other than a "coefficient"

 

[hasan_researc]

I am sorry, Arya svit-kon, but would you allow me and vela get to through the end of the problem, please?

 

[Arya_svit-kon]

yep sure, sorry :)

 

[vela]

OK, so

(b) possible outcomes of the second measurement = E_{1} and E_{2}
and probability of E_{1} = 2/3 and E_{2} = 1/3.

What do you think?

That's right.

 

[hasan_researc]

Does e^{-iE_{1}t/hcross} have no effect whatsoever on my calculation?

 

[hasan_researc]

Or e^{-iE_{2}t/hcross} for that matter?

 

[vela]

Sorry to go back to an earlier question, but when finding the probabilities of E1 and E2, how did you do it? I know you take "the modulus of an squared", and this will sound ridiculously lame (I am rubbish with quantum mechanics) but how do you get 2/3 and 1/3? because to take the modulus of E1 would be that of (sqrt(2)+1)all squared over 3 ? surely? although to be honest I don't really understand what "an" stands for other than a "coefficient"

Say a particle is in a state with wave function ψ(x). You can write the wave function ψ(x) as a linear combination of eigenfunctions ϕ_i of some observable P:

\psi(x) = a_1 \phi_1(x) + a_2 \phi_2(x) + \cdots + a_n \phi_n

(For simplicity, let's assume the corresponding eigenvalues λ_i are all distinct.) The probability of a measurement of P yielding the result λ_j is then just |a_j|². It's not all of the coefficients. It's just the one coefficient of the eigenfunction that corresponds to eigenvalue λ_j.

 

[vela]

Does e^{-iE_{1}t/hcross} have no effect whatsoever on my calculation?

Or e^{-iE_{2}t/hcross} for that matter?

How did you come up with 2/3 and 1/3 for your answers? The exponentials are part of the coefficient. If you ignored them, you did the calculation incorrectly.

 

[hasan_researc]

OK, let's be clear. The probability is given by the square of the modulus of the relevant coefficient.

And given what you have said right now, the coefficient is the exponential times the real number (associated with each u(x)) ??

Am I right?

If I am, then the exponential cancels out because we are taking the square of the modulus.

Am I right?

Just to speed things up a little, would you mind giving me a hint as to how I could go about doing part (c)?

 

[vela]

OK, let's be clear. The probability is given by the square of the modulus of the relevant coefficient.

And given what you have said right now, the coefficient is the exponential times the real number (associated with each u(x)) ??

Am I right?

If I am, then the exponential cancels out because we are taking the square of the modulus.

Am I right?

Yes, that's exactly right.

Just to speed things up a little, would you mind giving me a hint as to how I could go about doing part (c)?

Since now you're looking at a measurement of Q, you want to express the state in terms of the eigenstates of Q. Once you have that, again, you just look at the coefficients to get the probabilities.

 

[hasan_researc]

Whow! This looks so easy! I'll post my solution in a minute.

 

[hasan_researc]

So, this is what I've got to so far!

\psi(x, t) = \frac{ \sqrt{2}u_{1}(x)e^{-iE_{1}t/hcross} + u_{2}(x)e^{-iE_{2}t/hcross} }{ \sqrt{3} }

\phi_{1}(x) = \frac{\sqrt{2}u_{1}(x) + u_{2}(x)}{\sqrt{3}}

\phi_{1}(x) = \frac{u_{1}(x) - \sqrt{2}u_{2}(x)}{\sqrt{3}}

Solving the second and third equations simultaneously for u_{1}(x) and u_{2}(x), we have

u_{1} = \frac{ \sqrt{6}\phi_{1}(x) + \sqrt{3}\phi_{2}(x) }{ 3 }

u_{2} = \frac{ \sqrt{3}\phi_{1}(x) + \sqrt{6}\phi_{2}(x)}{ 3 },

which leads to

\psi (x , t) = \left( \frac{ \sqrt{6}e^{-iE_{1}t/hcross} + \sqrt{3}e^{-iE_{2}t/hcross}} {3} \right) \phi_{1} + \left( \frac{ \sqrt{3}e^{-iE_{1}t/hcross} + \sqrt{6}e^{-iE_{2}t/hcross}} {3} \right) \phi_{2}.

 

[hasan_researc]

What do you think?

 

[vela]

That's the right approach, but you dropped a sign somewhere when calculating u₂. Also, in ψ, you forgot the coefficients in front of u₁ and u₂.