# Measuring photon polarization and finding observable operator

An apparatus has these properties when measuring a polarized photon:
-whenever a linearly polarized photon at angle $$\vartheta$$ enters apparatus, it displays "2"
-whenever a linearly polarized photon at angle $$\frac{pi}{2}+\vartheta$$ enters apparatus, it displays "3"
-for all other polarizations other than above, it displays 2 or 3 with random probabilities.
1. Find the eigenvalues and eigenstates.
2.Find the matrices of of operator A in its eigenbasis and |H> |v> basis.

So for part 1, I believe the eigenvalues are 2 and 3. Then the eigenstates are |$$\vartheta$$><$$\vartheta$$| and |$$\frac{pi}{2}+\vartheta$$><$$\frac{pi}{2}+\vartheta$$|.
However, I am not sure how to find the eigenvalues and eigenstates for the third property of the apparatus. I probably has to do with the same eigenvalues but the states I am not sure of.
As for part 2, I do not understand it.
So any help would be appreciated!
Thank You
DoubleMint

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those aren't the eigenstates what you wrote down those are projection operators. The eigenstates are simply

$$\left| \theta \right\rangle , \left| \theta +\frac{\pi}{2} \right\rangle$$

whilst the third property is simply linear combinations of the two states

$$\Psi = \alpha \left| \theta \right\rangle + \alpha' \left| \theta +\frac{\pi}{2} \right\rangle$$

where

$$\left|\alpha \right| ^2 + \left|\alpha' \right| ^2 = 1$$

the matrix of an operator is given by

$$A_{ij} = \sum_{i,j=1}^{number of sates} \left\langle \psi_i \right| \hat{A} \left| \psi_j \right\rangle$$

Thanks for the reply sgd37. Now, how do i incorporate the third property into the matrix operator?

you only need the eigenbasis to define the matrix operator. Is that what you meant

you only need the eigenbasis to define the matrix operator. Is that what you meant
No, I mean how do you find the values associated with the linear combination of the third property in the operator matrix. I can find the values for projection operators of $$\vartheta$$ and $$\frac{pi}{2}\vartheta$$ which should be 2 and 3 respectively.

There are no values in the matrix associated with the linear combinations. And there is no way of predetermining the states

For the two states you can determine the operator matrix

$$A= \begin{pmatrix} 2&0\\ 0&3 \end{pmatrix}$$

for which the eigenvectors are

$$\left| \theta \right\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix} , \left| \theta +\frac{\pi}{2} \right\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}$$

the linear combinations just determine a general vector in $$R^2$$

I never knew that...interesting. Thanks alot sgd37!