Observables, Measurements and all that

[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

 

[stevendaryl]

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

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)

 

[tom.stoer]

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

This is correct.

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>.

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

 

[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

 

[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)

 

[spookyfw]

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