# Homework Help: Measurement in quantum mechanics

1. Sep 7, 2012

### Antepavolic

1. The problem statement, all variables and given/known data
Given a system initially in a state
$$| \psi \rangle = \frac{1}{\sqrt{6}} \left(|1\rangle + 2 |2\rangle + |3\rangle \right)$$
where ${|n\rangle}_{n=1}^{8}$form an ON basis.

If we perform a measurement of an observable corresponding to an operator
$$\hat{A}=|2\rangle\langle3|+|3\rangle\langle2|+|3\rangle 2 \langle3|.$$
What are the possible outcomes and probabilities of this measurement?

2. Relevant equations

3. The attempt at a solution
I know that a measurement will yield an eigenvalue corresponding to an eigenvector of the operator. The probability is given by $P(n)=|\langle n |\psi \rangle|^2$ if $|\psi\rangle$ is normalized. Now if I act with $\hat{A}$ on $|\psi\rangle$ I get:
$$\hat{A} |\psi\rangle = \frac{1}{\sqrt{6}} \left(| 2 \rangle + 4 |3\rangle \right).$$ So the only possible outcomes are eigenvalues to $| 2 \rangle$ and $| 3 \rangle$, right?

If so, how do I calculate the probabilities?

If I normalize $|\psi\rangle$ and then calculate $|\langle 2 | |\psi\rangle |^2$ and $|\langle 3 | |\psi\rangle |^2$ it seems I miss some information about the probable outcomes contained in the operator.

This is how far I got.

2. Sep 7, 2012

### Chopin

Since a measurement yields a state which is an eigenvector of the operator, your first step should be to figure out what those eigenvectors are. In other words, you should start by trying to diagonalize $A$.

3. Sep 7, 2012

### TSny

Hello, Antepavolic. Welcome to PF.

Right. So, you might want to find the eigenvalues and eigenvectors of the operator. [Edit: Chopin already pointed this out while I was working on my response.]

Yes, where $\langle n|$ corresponds to an eigenvector of $\hat{A}$ and $P(n)$ denotes the probability that your measurement yields the eigenvalue of $\hat{A}$ corresponding to the eigenvector $|n\rangle$

Not sure what you're saying here. As you said earlier, the only possible outcomes of the measurement are the eigenvalues of $\hat{A}$ and the measurement will throw the system into the state represented by the eigenvector of $\hat{A}$ corresponding to the eigenvalue that was obtained in the measurement.

Note that the expression $$\hat{A} |\psi\rangle = \frac{1}{\sqrt{6}} \left(| 2 \rangle + 4 |3\rangle \right)$$ does not correspond to a measurement of $\hat{A}$. A measurement of $\hat{A}$ will collapse the wavefunction to one of the eigenvectors of $\hat{A}$. The vector on the right hand side of $\hat{A} |\psi\rangle$ is not an eigenvector of $\hat{A}$.

Don't fall into the trap of thinking that a "measurement of an observable" is represented mathematically by operating on the wavefunction with the operator corresponding to the observable. The result of operating on the initial wavefunction by the operator does not yield the wavefunction corresponding to the result of the measurement.

As you said earlier, the probabilities are given by $P(n)=|\langle n |\psi \rangle|^2$.
This won't give you the answer for the probabilities of the outcomes for the measurement of $\hat{A}$ because $|2\rangle$ and $|3\rangle$ are not eigenvectors of $\hat{A}$.

4. Sep 8, 2012

### Antepavolic

I think I got it.

I let $\hat{A}$ be represented by a matrix in the given ON basis. This gave me a 8x8 matrix with mostly zeros. I then calculated the eigenvalues to:

$$\lambda_1 = 0, \ \lambda_2 = \frac{1}{1+\sqrt{2}}, \ \lambda_3 = \frac{1}{1-\sqrt{2}}, \ \lambda_i = 0 (i=4,5,6,7,8)$$

I calculated the eigenvectors and then the probabilities according to the recipe $P(\lambda_i)= |\langle a_n | \psi \rangle|^2$ where $a_n$ denotes the normalized eigenvectors.

At last I checked that the probabilities added up to one.

Thank you for helping out.