# What is the state vector after measuring

1. Feb 21, 2013

### aaaa202

What is the state vector of the system after measuring a degenerate eigenvalue?

Surely there are infinitely many vectors of norm 1 that it could be since the degenerate eigenvector span a 2d subspace.

Say that we e.g. have a system where the operators eigenvalues are degenerate such that:

(1,0), (0,1) both represent eigenvectors of same eigenvalue. How can we ever know what state the particle is in after measurement?

2. Feb 22, 2013

### Simon Bridge

This is correct - the measurement establishes the eigenvalue, not the vector.
The result will be a linear superposition of the degenerate states depending on the nature of the measurements.
i.e. if the system is a particle in a box, and a measurement of position is taken, and the particle position-state is found to be $S=\langle x|\psi \rangle$ then that position could have resulted from any number of possible energy eigenstates ... so it is infinitely degenerate. The possible results of future measurements can be predicted by the time evolution of the superposition of eigenstates corresponding to the outcome of the measurement.

In this case it is usually more convenient to use the density matrix formulation.

See. http://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics#Density_matrix_formulation

3. Feb 22, 2013

### vanhees71

Of course, this depends on the measurement apparatus. Often you simply destroy the object you measure (like detecting a photon, and it gets absorbed). Then you don't want to know the state after the measurement, and it would be a pretty hard task to give it, because it's a complicated excitated state of the macroscopic measurement apparatus that has just absorbed a photon.

If you have an ideas von Neumann measurement (also known as projective measurement), like in an Stern-Gerlach experiment, where you split a beam of atoms with non-zero spin into a set of beams ordered according to the spin-magnetic quantum number by running the atoms through an inhomogeneous magnetic field. Then you absorbed all partial beams except of one, giving you a beam with a definite spin-magnetic quantum number $\sigma_z$.

If the spin part of the single atoms in the original beam are described by the (normalized) state vector $|\psi \rangle$ then after this procedure, the partial beam is described by the statevector
$$\tilde{\psi}'= |\sigma_z \rangle \langle \sigma_z|\psi \rangle.$$
The intensity of the new beam then is
$$\langle \tilde{\psi}'|\tilde{\psi}' \rangle.$$
Then you normalized the state vector again to make experiments with the ideally polarized beam:
$$|\psi' \rangle=\frac{1}{\|\tilde{\psi}' \|} |\tilde{\psi}' \rangle.$$

4. Feb 22, 2013

### aaaa202

well maybe I don't understand or you missed my point. Let me make it more clear with an example. Suppose there exists an observable represented by a 2x2 matrix with degenerate eigenvalues,
Now it is clear that any vector in R2 is an eigenvector to this observable, because all vectors in R2 can be represented by a linear combination of the two degenerate eigenvalues. So if you measured an eigenvalue corresponding to this observable how would you ever know which state you had collapsed the state vector onto?

5. Feb 22, 2013

### Simon Bridge

What would a "measurement" involve here?
Show me the math.

6. Feb 22, 2013

### Fredrik

Staff Emeritus
That would be a matrix of the form λI, where I is the identity matrix. This matrix represents a measuring device such that every measurement has the result λ. Such a device doesn't even have to interact with the system. Basically, it's a post-it note with the number λ written on it.

You wouldn't.

7. Feb 23, 2013

### jk22

if you have a measurement operator with degenerate eigenvalue the final state is given as in the usual case : it is the normalized projection of the initial state onto the eigenspace.
However if you had no measurement operator, like a one side measurement of an entangled bipartite state, then the final state is unknow and given by some vector in this space.

8. Feb 23, 2013

### Staff: Mentor

I don't have a different/better answer than some of the other posters have already given you, but I will suggest that you look into Ballentine's distinction between preparation and measurement - much easier to speak precisely about states "before" and "after". I learned this from a paper copy of his book, but I'm pretty sure that someone here can find a pointer to something online.