# The interference term. Confused.

1. Jun 27, 2011

### LostConjugate

Can someone explain what the difference is between the expectation value of an observable in a pure vs a mixed state. The equations are identical. For example with spin up and down $$|A|^2<S_u|O|S_u> + |B|^2<S_d|O|S_d>$$

2. Jun 27, 2011

### Bill_K

What you've quoted is <O> for the case where the system is in an incoherent mixture of |Su> and |Sd>. This would be a classical mixture, like heads and tails on a coin, and is represented by a density matrix rather than a wavefunction.

More typically what we mean by a mixed state is a coherent mixture, |S> = A |Su> + B |Sd>, in which case the expectation value <S|O|S> will need to include interference terms A*B <Su|O|Sd> and B*A <Sd|O|Su>.

3. Jun 27, 2011

### LostConjugate

If you could quickly reference page 358 of this pdf http://www.physics.sfsu.edu/~greensit/book.pdf [Broken]

It shows that your equation is equivalent to mine...

Last edited by a moderator: May 5, 2017
4. Jun 27, 2011

### kith

That's true only if your states are eigenstates of O (so the interference terms vanish).

This means, you can't distinguish between a coherent superposition and an incoherent mixture by measuring in one basis alone. But you don't get identical expectation values in all bases!

Consider a superposition of 50:50 spin up and spin down along a given axis. If you change your measurement axis (corresponding to another basis) you get other ratios. In the case of incoherent mixing, the ratio is 50:50 regardless of basis. You can verify this by an easy calculation of the corresponding density matrices.

5. Jun 27, 2011

### LostConjugate

What are the terms that vanish. If the system is mixed or pure it does not have any extra terms.

Edit: I see the term now, I looked right over it, thanks!

Last edited: Jun 27, 2011