# Should spacial functions be involved when calculating <Sx>?

• Haorong Wu
In summary, Perok advised that the spatial wave function is not relevant for the solution to the Spin Problem. Solution 2 is given by the book, which has errors, so the student is unsure whether they are right or the book is right. Solution 1 uses the spin coefficients to calculate the spin probabilities. When solving the Spin Problem using Dirac notation, it becomes more clear that the spatial wave function is not relevant.
Haorong Wu
Homework Statement
Given a state of a electron, ##\psi = \begin{pmatrix} \psi_{+} \\ \psi _{-} \end{pmatrix} =R \left ( r \right ) \begin{pmatrix} \sqrt {\frac 3 5} Y_0^0 + \sqrt {\frac 1 {10}} Y_1^1 +\sqrt {\frac 1 {10}} Y_1^-1 \\ \sqrt {\frac 1 5} Y_1^0 \end{pmatrix}##, what is the expectation of ##\left < S_x \right >## ?
Relevant Equations
None
I have two different solutions, and I do not know which one is correct and why the other one is wrong.

Solution 1.
In the ##L_z## space, the spin state is ##\begin{pmatrix} \sqrt { \frac 4 5} \\ \sqrt { \frac 1 5} \end{pmatrix}##, and ##S_x=\frac \hbar 2 \begin{pmatrix} 0& 1 \\ 1& 0 \end{pmatrix}##, so ##\left < S_x \right >=\begin{pmatrix} \sqrt { \frac 4 5} & \sqrt { \frac 1 5} \end{pmatrix} \frac \hbar 2 \begin{pmatrix} 0& 1 \\ 1& 0 \end{pmatrix} \begin{pmatrix} \sqrt { \frac 4 5} \\ \sqrt { \frac 1 5} \end{pmatrix}=\frac 2 5 \hbar##.

Solution 2.
##\left < S_x \right >=\int \begin{pmatrix} \psi_{+}^{*} & \psi_{-}^{*} \end{pmatrix} \frac \hbar 2 \begin{pmatrix} 0& 1 \\ 1& 0 \end{pmatrix} \begin{pmatrix} \psi_{+} \\ \psi_{-} \end{pmatrix} dr =0##.

I guess the problem is that I have not make clear whether the spatial functions should be involved in the calculations. Should the orbital momentums affect the calculation of spins?

Thanks!

In your second method, would you always get 0?

Ps my phone won't render all of what you've written. In general the spatial wave function is not relevant. You can see this by expressing the spin operator on the combined space as the tensor product of the identity operator on position space and the usual spin operator.

Sorry, it's difficult to type formulas as I'm on a bumpy train!

Haorong Wu
PeroK said:
Ps my phone won't render all of what you've written. In general the spatial wave function is not relevant. You can see this by expressing the spin operator on the combined space as the tensor product of the identity operator on position space and the usual spin operator.

Sorry, it's difficult to type formulas as I'm on a bumpy train!
Thanks, Perok.

In the second method, ##\begin{pmatrix} 0 & 1 \\ 1& 0 \end{pmatrix} \cdot \begin{pmatrix} \psi_+ \\ \psi_- \end{pmatrix}=\begin{pmatrix} \psi_- \\ \psi_+\end{pmatrix}##, and all the ##Y_l^m## in ##\psi_+## and ##\psi_-## are different so that they are orthogonal, so ##\begin{pmatrix} \psi_+^* & \psi_-^* \end{pmatrix} \cdot \begin{pmatrix} \psi_- \\ \psi_+\end{pmatrix}=0##.

In fact, I think that spatial wave function is not relevent, too. However, the second solution is given by the book, which has many errors, so I am not sure whether I am wrong or the book is wrong.

I will try the tesor product later. Again, thanks for your advice.

Haorong Wu said:
Thanks, Perok.

In the second method, ##\begin{pmatrix} 0 & 1 \\ 1& 0 \end{pmatrix} \cdot \begin{pmatrix} \psi_+ \\ \psi_- \end{pmatrix}=\begin{pmatrix} \psi_- \\ \psi_+\end{pmatrix}##, and all the ##Y_l^m## in ##\psi_+## and ##\psi_-## are different so that they are orthogonal, so ##\begin{pmatrix} \psi_+^* & \psi_-^* \end{pmatrix} \cdot \begin{pmatrix} \psi_- \\ \psi_+\end{pmatrix}=0##.

In fact, I think that spatial wave function is not relevent, too. However, the second solution is given by the book, which has many errors, so I am not sure whether I am wrong or the book is wrong.

I will try the tesor product later. Again, thanks for your advice.

Edit: Ignore this!

My understanding of that notation is that we are using the coefficients of spin in the z-basis to express the full wave function. This includes the normalisation condition. In that case it's clear you can get the spin probabilities normally.

The spatial wave function is then a superposition with the coefficients likewise determined by the spin coefficients. Although, as in your example, each of the spatial wave functions themselves can be further broken down into a superposition of spatial wave functions. This does not affect the overall spin coefficients or probabilities.

Last edited:
PeroK said:
The spatial wave function is then a superposition with the coefficients likewise determined by the spin coefficients. Although, as in your example, each of the spatial wave functions themselves can be further broken down into a superposition of spatial wave functions. This does not affect the overall spin coefficients or probabilities.
I disagree. In the ##\psi## state, the spin state is position dependent. One therefore cannot talk about "overall" coefficients or probabilities. The full wave function has to be considered and solution 2 is the correct one.

This becomes more obvious when the problem is tackled using Dirac notation.

Haorong Wu
DrClaude said:
I disagree. In the ##\psi## state, the spin state is position dependent. One therefore cannot talk about "overall" coefficients or probabilities. The full wave function has to be considered and solution 2 is the correct one.

This becomes more obvious when the problem is tackled using Dirac notation.
Yes, I'm afraid I was guessing about what was on the right hand side of the screen.

It's obviously not what I was guessing!

Apologies to @Haorong Wu.

PeroK said:
Yes, I'm afraid I was guessing about what was on the right hand side of the screen.

It's obviously not what I was guessing!

Apologies to @Haorong Wu.
Thanks, PeroK and DrClaude.

I will keep tracking the spatial wave function when I encounter a similar situation.

Thanks again.

## 1. What is the meaning of "spacial functions" in this context?

In this context, "spacial functions" refer to mathematical functions that describe the position and movement of objects in physical space. This can include functions such as position, velocity, and acceleration.

## 2. Why is it important to consider spacial functions when calculating ?

Spacial functions are important to consider when calculating because they provide information about the location and movement of objects, which can affect the value of . Ignoring spacial functions could lead to inaccurate calculations and results.

## 3. Can spacial functions be ignored in certain situations when calculating ?

It is generally not recommended to ignore spacial functions when calculating . However, in some cases where the object's movement is negligible or has no impact on the measurement of , spacial functions may not need to be considered.

## 4. Are there specific spacial functions that are more important to consider when calculating ?

The specific spacial functions to consider when calculating may vary depending on the situation. However, functions related to the object's position and movement in the direction of may have a greater impact on the value of . It is important to carefully consider all relevant spacial functions in each individual case.

## 5. How can spacial functions be incorporated into the calculation of ?

Spacial functions can be incorporated into the calculation of by using mathematical equations that involve the object's position, velocity, and acceleration in the direction of . These functions can be integrated or differentiated to determine the value of at a specific point in time.

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