Probability of Measuring ##L_x = 0## for a Given ##\psi##

[t0pquark]

Homework Statement

Compute the probability that a measurement of $L_x$ will give zero for an angular momentum particle (1) with $\vert \psi \rangle = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix}$.

Relevant Equations

In a 3D Hilbert space, for spin 1, $L_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$.

The probability that a measurement of $L_x$ will give zero for a given $\psi$ should be $\vert \langle L_x = 0 \vert \psi \rangle \vert ^2$, I think.
I found the eigenvalues of $L_x$ to be $\lambda = -1, 0, 1$. Since it asks for the probability that the measurement will give zero, I think I should be looking at the eigenvector that corresponds to zero, which is $\begin{bmatrix} \frac{1}{\sqrt 2} \\ 0 \\ \frac{-1}{\sqrt 2} \end{bmatrix}$.
I am not sure where to go from here?

 

[DrClaude]

Simply calculate $| \langle L_x = 0 | \psi \rangle |^2$ in vector form.

 

[PeroK]

Homework Statement: Compute the probability that a measurement of $L_x$ will give zero for an angular momentum particle (1) with $\vert \psi \rangle = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix}$.
Homework Equations: In a 3D Hilbert space, for spin 1, $L_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$.

The probability that a measurement of $L_x$ will give zero for a given $\psi$ should be $\vert \langle L_x = 0 \vert \psi \rangle \vert ^2$, I think.
I found the eigenvalues of $L_x$ to be $\lambda = -1, 0, 1$. Since it asks for the probability that the measurement will give zero, I think I should be looking at the eigenvector that corresponds to zero, which is $\begin{bmatrix} \frac{1}{\sqrt 2} \\ 0 \\ \frac{-1}{\sqrt 2} \end{bmatrix}$.
I am not sure where to go from here?

You need to think about what you are doing. You have a state expressed in one basis ( the z-basis). And you really want that state expressed in another basis (the x-basis).

So, you could do all of the work to transform your state to the x-basis and then you could simply read off the probabilities of getting $-1, 0, 1$ for a measurement of $L_x$: just take the modulus squared of each component.

But, you are only asked for the probability of getting $L_x = 0$. So, you only need the zeroth component in the x-basis. So, you don't have to do all the work. You only need to find one component.

Now, if you think about linear algebra, the components of any vector $\psi$ in any basis are given by $\langle b_n | \psi \rangle$, where $b_n$ is nth basis vector.

In this case, you want the zeroth component of $\psi$ in the x-basis. As you have calculated the zeroth x-basis vector, you can just take the inner product of this with $\psi$. The probability, as above, is the modulus squared of this component.

 

[t0pquark]

That explanation helps!

So then I have $( \begin{bmatrix} \frac{1}{ \sqrt 2} \\ 0 \\ \frac{-1}{ \sqrt 2} \end{bmatrix} \cdot \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix} )^2 = (\frac{2}{3 \sqrt 2} + \frac{2}{3 \sqrt 2})^2 = \frac{8}{9}$, right?

 

[DrClaude]

That explanation helps!

So then I have $( \begin{bmatrix} \frac{1}{ \sqrt 2} \\ 0 \\ \frac{-1}{ \sqrt 2} \end{bmatrix} \cdot \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix} )^2 = (\frac{2}{3 \sqrt 2} + \frac{2}{3 \sqrt 2})^2 = \frac{8}{9}$, right?

The left vector should be a row vector, but otherwise it is correct.