# Observables, Measurements and all that

1. Mar 2, 2013

### spookyfw

Hi Folks,

I somehow cannot get the difference and have to admit that I am left confused.

For a probability of measuring m with the operator M on state $\Psi_i$
$p(m|i) = <\Psi_i| M^{+}_m M_m |\Psi_i> = <\Psi_i| M_m |\Psi_i>$.

The average of an observable is defined as $<O> = <\Psi_i| O |\Psi_i>$.

So the measurement by M gives me a probability, the measurement with the observable an expectation value? Okay..the observables will be hermitean, the only thing I know about the measurement matrix is, that is not unitary - otherwise $M^{+}_m M_m$ would be equal to the unity matrix.

Is the difference that an observable doesn't change the system, but a measurement when projective projects the system into one of the states?

One last question: What is the reasoning for defining a measurement like $M^{+}_m M_m$ and not by M alone directly?

Thank you so much in advance..I hope the above somehow makes sense ;).
Steffen

2. Mar 2, 2013

### stevendaryl

Staff Emeritus
Let $\vert \Phi_m \rangle$ be a complete set of eigenstates of operator $M$: (for simplicity, let's assume that the eigenstates are nondegenerate; that is, there can't be more than one state with eigenvalue $m$)

1. $M \vert \Phi_m \rangle = m \vert \Phi_m \rangle$
2. $\langle \Phi_m \vert \Phi_m \rangle = 1$
3. $\langle \Phi_m' \vert \Phi_m \rangle = 0$, if $m' \neq m$

The idea behind a "projection operator" $M_m$ is this:

$M_m \vert \Phi_m \rangle = \vert \Phi_m \rangle$
$M_m \vert \Phi_m' \rangle = 0$, if $m' \neq m$.

So $M_m$ filters out any eigenstates that do not have eigenvalue $m$.

If you have such a projection operator, then the action of $M_m$ on an arbitrary wave function $\vert \Psi_i \rangle$ can be computed this way:

• Write $\vert \Psi_i \rangle$ as a superposition of eigenstates of $M$:

$\vert \Psi_i \rangle = \sum_{m'} C_{m'} \vert \Phi_{m'} \rangle$

where $C_{m'}$ is just a coefficient. Note that $C_{m'}$ is the amplitude for state $\Psi_i$ to have eigenvalue $m$. The probability is the square: $P(m'\vert i) = \vert C_{m'} \vert^2$.
• Now apply the projection operator $M_m$:
$M_m \vert \Psi_i \rangle = \sum_m' C_m' M_m \vert \Phi_m' \rangle$
• By the assumed properties of $M_m$, all the terms in the sum vanish except one, leaving:
$M_m \vert \Psi_i \rangle = C_m \vert \Phi_m \rangle$
• Now form the quantity $\vert M_m \vert \Psi_i \rangle \vert^2$:
$\vert M_m \vert \Psi_i \rangle \vert^2 = \langle \Psi_i M_m^\dagger M_m \vert \Psi_i \rangle = \langle \Phi_m C_m^* C_m \vert \Phi_m\rangle = \vert C_m \vert^2 = P(m \vert i)$
• So we conclude: $\langle \Psi_i M_m^\dagger M_m \vert \Psi_i \rangle = P(m \vert i)$

3. Mar 3, 2013

### tom.stoer

This is correct.

I guess what you mean is that M is a projector on an eigenstate with eigenvalue m (w/o degeneration).

I would use the letter M for an observable associated with m with appropriate (non-degenerate) eigenstates

$$M |m\rangle| = m |m\rangle$$

and I would use P to indicate that we have an projector. So we have a clear distinction between the observable M w.r.t. which we define the eigenvalues and eigenstates and the projectors which project an arbitrary state to an eigenstate.

That means

$$p(m|i) = | \langle m|\Psi_i\rangle|^2 = \langle\Psi_i|m\rangle\langle m|\Psi_i\rangle = \langle\Psi_i|P_m|\Psi_i\rangle = \langle\Psi_i|P_m^\dagger\,P_m|\Psi_i\rangle$$

where I use

$$|m\rangle\langle m| = P_m$$
$$P_m = P_m^2$$
$$P_m = P_m^\dagger$$

Last edited: Mar 3, 2013
4. Mar 3, 2013

### spookyfw

Stevendaryl and tom.stoer! Thank you very much for your replies. I think I got it finally, one question that directly follows: is it then right to say that:

$<M> = \Sigma p(m|i)m$

and hence $M = \Sigma P_m m$

5. Mar 3, 2013

### tom.stoer

Yes. For the observable M you can write

$$M = \sum_m m\,|m\rangle\langle m|= \sum_m m\,P_m$$

Everything else follows from this spectral representation (where I have used the assumptions of descrete and non-degenerate eigenvalues)

6. Mar 3, 2013

### spookyfw

Ah..perfect. So the loop is closed :). Thanks once again!