Are Probabilities the Same for Operator A in Different Basis?

[Niles]

Homework Statement

Hi all.

I have the following state at t=0 in a 3D Hilbert-space (it is in the eigenspace of the 3x3 Hamiltonian):

$$\left| \psi \right\rangle = \frac{1}{{\sqrt 2 }}\left| \psi \right\rangle _1 + \frac{1}{2}\left| \psi \right\rangle _2 + \frac{1}{2}\left| \psi \right\rangle _3.$$

Now I have an operator representing an observal given by:

$$\hat A = \left( {\begin{array}{*{20}c}<br /> 1 & 0 & 0 \\<br /> 0 & 0 & 1 \\<br /> 0 & 1 & 0 \\<br /> \end{array}} \right)$$

I have to find the possible eigenvalues of A and the corresponding probabilities.

The Attempt at a Solution

The possible eigenvalues of A are easy. I am more concerned about the probabilities. I reasoned that they are the same, because the above state at t=0 is independent of the Hilbert space in is written in. So it will look the same if I write it in the eigenspace of A, but the unit-vectors (i.e. the possible states) are now different.

So my attempt: The probabilities are the same, i.e. 1/2, 1/4 and 1/4. Can you confirm this?

Thanks in advance.


Niles.

 

[Niles]

Ok, I think I got it now.

My above suggestion is not correct. I think I have to find the coefficients by writing the state |psi> in the basis of A by the following method:

$$a_{cofficient\,\,1} = \left\langle {{\psi _{a\,\,1} }}<br /> \mathrel{\left | {\vphantom {{\psi _{a\,\,1} } \psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\psi } \right\rangle = \sum\limits_i {\left\langle {{\psi _{a\,\,1} }}<br /> \mathrel{\left | {\vphantom {{\psi _{a\,\,1} } {\psi _i }}}<br /> \right. \kern-\nulldelimiterspace}<br /> {{\psi _i }} \right\rangle \left\langle {{\psi _i }}<br /> \mathrel{\left | {\vphantom {{\psi _i } \psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\psi } \right\rangle },$$

where psi_i are the eigenfunctions of the Hamiltonian and psi_{a 1} are the eigenfunctions of A. The above method gives me the first coefficient for the representation of |psi> in the basis of A.

Is this correct?