What is the Probability of Measuring 2a in State |\varphi (t)\rangle\right?

[Denver Dang]

Homework Statement

Find the probability, $P_{2a}(t)$, that a measurement of the quantity $A$ in
the state $|\varphi (t)\rangle\right$ will yield the value $2a$.

Homework Equations

$$\hat{A}|1\rangle\right = a(|1\rangle\right - i|2\rangle\right$$
$$\hat{A}|2\rangle\right = a(i|1\rangle\right + |2\rangle\right$$
$$\hat{A}|3\rangle\right = -2a(|3\rangle\right$$

$$A = \[ \left( \begin{array}{ccc}<br /> a & ia & 0 \\<br /> -ia & a & 0 \\<br /> 0 & 0 & -2a\end{array} \right)\]$$

$$|\varphi (t)\rangle\right = \[ \left( \begin{array}{ccc}<br /> cos(wt) \\<br /> 0 \\<br /> -isin(wt) \end{array} \right)\]$$

The Attempt at a Solution

Well, I kinda suck at finding these probabilities. So I'm not sure what to do, since it asks for $2a$. Is it just:
$$P(2a) = \left|\langle\psi_j|\Psi\rangle\right|^2,$$
where $\psi_j = \varphi$ and $\Psi = A|3\rangle\right$, or am I just not getting it ?


Regards.

 

[nickjer]

You need to find the eigenvectors and eigenvalues of the matrix, A, in terms of the basis vectors you are given. So basically you need to diagonalize the matrix you are given. This will give you 3 eigenvectors that are superpositions of the states: $|1\rangle,|2\rangle,|3\rangle$.

Once you have the eigenvectors of 'A': $|1'\rangle, |2'\rangle, |3'\rangle$ (all of which you need to properly normalize) and their corresponding eigenvectors. Then you will choose the eigenvector $|u'\rangle$ with the eigenvalue, +2a. Since this will be the superposition state the wavefunction will collapse to once you measure the property A and find its value +2a.

Finally, as you said before, the probability of finding this superposition state is:

$$P(2a) = \frac{\left|\langle u'|\varphi(t)\rangle\right|^2}{\left|\langle \varphi(t)|\varphi(t) \rangle \right|^2}$$

 

[Denver Dang]

So I get the eigenvectors to be:

$[-i,1,0]$, $[0,0,1]$ and $[i,1,0]$.

Normalized they will become:

$[\frac{-i}{\sqrt{2}}, 0, 1]$, $[\frac{1}{\sqrt{2},0]$, and the last one can't be normalized, or am I wrong ?

The eigenvalues is:

$[0, -2a, 2a],$

So I need to use the 3rd eigenvector ?

And using the formula, I get that the probability must be:

$$P_{2a} = \frac{cos^{2}(wt)}{4},$$
or am I way off ?Regards, and sorry for the late reply.

 

[nickjer]

You have the correct eigenvectors, but the normalized versions of them are wrong. You are correct about the eigenvalues and choosing the third eigenvector. But your probability is wrong. Double check the denominator in that probability.

 

[Denver Dang]

Think I screwed up the normalized eigenvectors. It should be:

$[\frac{-i}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0]$, $[0,0,1]$, and then again, the 3rd and last, I don't think I can normalize, since $\sqrt{i^{2} + 1^{1} + 0} = \sqrt{0}$

And I think I used the 3rd eigenvector instead of $\varphi(t)$ in the denominator, so that's why I got it wrong. It should of course be:

$$P_{2a} = cos^{2}(wt)$$

Or maybe divided by 2 if I can normalize the eigenvector as the 1st one ?

 

[nickjer]

The third vector should normalized the same as the first.

You want to multiply the vector by some normalization constant and solve for that constant. Remember that the magnitude of a vector is the dot product of itself with its complex conjugate. I believe you are leaving off the complex conjugate in the multiplication.

 

[Denver Dang]

Yup... I see now :)

Once again, thank you very much.