Should spacial functions be involved when calculating <Sx>?

[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!

 

[PeroK]

In your second method, would you always get 0?

 

[PeroK]

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]

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.

 

[PeroK]

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.

 

[DrClaude]

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.

 

[PeroK]

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.

 

[Haorong Wu]

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.