# Expectation value via trace

1. Jun 11, 2012

### sunrah

1. The problem statement, all variables and given/known data
given
$\mid \psi \rangle = \frac{1}{\sqrt{2}} (\mid1\rangle + \mid2\rangle )$
where $\mid1\rangle, \mid2\rangle$ are orthonormal
calculate
i)density operator
ii) $\langle A \rangle$ where A is an observable
2. Relevant equations

3. The attempt at a solution
i) $\rho = \frac{1}{2} (\mid1\rangle\langle1\mid + \mid2\rangle\langle2\mid)$

ii) $\langle A \rangle = \frac{1}{2}\langle1\mid A\mid1\rangle + \frac{1}{2}\langle2\mid A\mid2\rangle + \frac{1}{2}\langle2\mid A\mid1\rangle + \frac{1}{2}\langle1\mid A\mid2\rangle$
i guess $\frac{1}{2}\langle2\mid A\mid1\rangle + \frac{1}{2}\langle1\mid A\mid2\rangle$ = 0 (because first 2 terms are each half the expected value but surely depends on operator therefore not necessarily zero??)

also I would like to use the trace to solve this:
$\langle A \rangle = tr[\rho A] = tr[\frac{1}{2}\mid1\rangle \langle1\mid A + \frac{1}{2}\mid2\rangle \langle2\mid A]$ but what does this mean?

Last edited: Jun 11, 2012
2. Jun 11, 2012

### vela

Staff Emeritus
This is the density matrix for a mixed state, where half the particles are in state $\vert 1 \rangle$ and half are in state $\vert 2 \rangle$. The state you were given is a pure state.

You can't really say anything about those matrix elements since you don't know anything about $\hat{A}$ except that it's an observable.

Say you have some basis and you find the matrix representing $\rho\hat{A}$. The trace is simply the sum of the diagonal elements. How would you write that in Dirac notation?

3. Jun 12, 2012

### sunrah

thanks, I have since worked out that the density matrix I gave is not complete (I assumed that the outer product of two orthogonal vectors would be zero!!)