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

• t0pquark
In summary: So the probability that a measurement of ##L_x## will give zero for the given state is ##\frac{8}{9}##. In summary, the probability that a measurement of ##L_x## will give zero for the given state is ##\frac{8}{9}##.
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?

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

t0pquark said:
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.

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?

PeroK
t0pquark said:
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.

## What is the meaning of "Probability of Measuring ##L_x = 0## for a Given ##\psi##"?

The probability of measuring ##L_x = 0## for a given state vector ##\psi## is the likelihood of obtaining a measurement result of zero for the x-component of angular momentum when the system is in the state described by ##\psi##. This is a measure of the uncertainty or randomness associated with the measurement.

## How is the probability of measuring ##L_x = 0## calculated?

The probability of measuring ##L_x = 0## for a given state vector ##\psi## can be calculated using the Born rule, which states that the probability of obtaining a particular measurement result is equal to the square of the amplitude of that measurement result in the state vector. In this case, the amplitude of ##L_x = 0## is the coefficient of the corresponding term in the wavefunction expansion.

## What factors affect the probability of measuring ##L_x = 0##?

The probability of measuring ##L_x = 0## for a given state vector ##\psi## is influenced by the relative contributions of different angular momentum states in the wavefunction expansion, as well as any external fields or interactions that may be present. Additionally, the orientation of the measurement apparatus with respect to the system can also play a role.

## How does the probability of measuring ##L_x = 0## change for different states of the system?

The probability of measuring ##L_x = 0## can vary greatly for different states of the system. For example, a state with a high amplitude for ##L_x = 0## in the wavefunction expansion will have a higher probability of measuring ##L_x = 0## compared to a state with a low amplitude for ##L_x = 0##. Additionally, changing the state of the system through time evolution or measurement collapses can also alter the probability of measuring ##L_x = 0##.

## How does the probability of measuring ##L_x = 0## relate to other measurements of angular momentum?

The probability of measuring ##L_x = 0## is just one aspect of the overall measurement of angular momentum. Other measurements, such as those for ##L_y## or ##L_z##, may also have their own probabilities associated with them. In some cases, the probabilities for different measurements may be related through the principles of quantum mechanics, such as the Heisenberg uncertainty principle.

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