QM:Expectation values and calculating probabilities

[Wavefunction]

Homework Statement

An operator \mathbf{A}, corresponding to a physical quantity \alpha, has two normalized eigenfunctions \psi_1(x)\quad \text{and}\quad \psi_2(x), with eigenvalues a_1 \quad\text{and}\quad a_2. An operator \mathbf{B}, corresponding to another physical quantity \beta, has normalized eigenfunctions \phi_1(x)\quad \text{and}\quad \phi_2(x), with eigenvalues b_1 \quad\text{and}\quad b_2. \alpha is measured and the value a_1 is obtained. If \beta is then measured and then \alpha again, show that the probability of obtaining a_1 a second time is \frac{97}{169}.

Homework Equations

The eigenfunctions are related via:

\psi_1 = \frac{(2 \phi_1+3 \phi_2)}{\sqrt{13}}
\psi_2 = \frac{(3 \phi_1-2 \phi_2)}{\sqrt{13}}

The Attempt at a Solution

Okay now I know I can represent |\psi\rangle by:
|\psi\rangle = \frac{1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}|\phi\rangle

I also know that initially:

\mathbf{A}|\psi\rangle = a_1|\psi\rangle = \frac{a_1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}|\phi\rangle which I can then bra through by \langle \psi| in order to get a_1 \langle \psi |\psi \rangle

Here's where I'm stuck, but I think maybe I should repeat the above process with the respective operators to get something like \langle\mathbf{A}_{\alpha}\rangle \langle\mathbf{B}_{\beta}\rangle \langle\mathbf{A}_{\alpha}\rangle

However, I'm unsure because I'm not very familiar with QM and I'm trying to prepare for the class before it begins this fall. Thanks for your help everyone.

 

[strangerep]

It's a lot simpler than that -- I suspect you're over-thinking the problem.

I'm not sure how much I should give away in a first response, so I'll just offer an initial hint:

What is the probability that a state prepared in state $\psi_i$ will be detected as state $\phi_j$ ?
In other words, what is $P(\phi_j|\psi_i)$ (which is read as "probability of $\phi_j$, given $\psi_i$) ?

BTW, which textbook(s) are you working from? If you have Ballentine, then you might be able to deduce the answer to my question from his eq(2.28).

 

[Wavefunction]

It's a lot simpler than that -- I suspect you're over-thinking the problem.

I'm not sure how much I should give away in a first response, so I'll just offer an initial hint:

What is the probability that a state prepared in state $\psi_i$ will be detected as state $\phi_j$ ?
In other words, what is $P(\phi_j|\psi_i)$ (which is read as "probability of $\phi_j$, given $\psi_i$) ?

BTW, which textbook(s) are you working from? If you have Ballentine, then you might be able to deduce the answer to my question from his eq(2.28).

Well the text I have is Modern Quantum Mechanics 2nd Edition by J.J. Sakurai; however, I'm just picking random problems from different sources to get a feel for QM. This particular problem is from here: http://farside.ph.utexas.edu/teaching/qmech/Quantum/node44.html

In regards to the problem:P(\phi_j|\psi_i) = |\langle \phi_j|\psi_i\rangle|^2 correct? So then the probability of measuring b_1 when the system is in the state \psi_1 is |\langle \phi_1|\psi_1\rangle|^2=\frac{4}{13}

 

[CAF123]

In regards to the problem:P(\phi_j|\psi_i) = |\langle \phi_j|\psi_i\rangle|^2 correct? So then the probability of measuring b_1 when the system is in the state \psi_1 is |\langle \phi_1|\psi_1\rangle|^2=\frac{4}{13}

Correct, so having measured $b_1$, the system is now in the state $\phi_1$. Then a third measurement is made, this time for A. What is the probability that, for your system in the state $\phi_1$, that you measure $a_1$?

 

[Wavefunction]

Correct, so having measured $b_1$, the system is now in the state $\phi_1$. Then a third measurement is made, this time for A. What is the probability that, for your system in the state $\phi_1$, that you measure $a_1$?

Ah okay that would be |\langle \psi_1|\phi_1\rangle|^2 =\frac{4}{13}

 

[CAF123]

Ah okay that would be |\langle \psi_1|\phi_1\rangle|^2 =\frac{4}{13}

Correct, can you finish the problem?

 

[Wavefunction]

Correct, can you finish the problem?

I think so because I think now I need to multiply the two probabilities together to get \frac{16}{169} Then I need to repeat the process for the state \phi_2:

|\langle \phi_2|\psi_1\rangle|^2 = \frac{9}{13}

Measuring \mathbf{A} again:

|\langle \psi_1|\phi_2\rangle|^2 = \frac{9}{13}

Multiplying together: \frac{81}{169}

Now adding the two together yields \frac{16+81}{169} = \frac{97}{169}

 

[CAF123]

Yes, looks fine.