# Schrodinger solution spin half particles

## Homework Statement

The evolution of a particular spin-half particle is given by the Hamiltonian $\hat{H} = \omega\hat{S}_{z},$ where $\hat{S}_{z}$ is the spin projection operator.
a) Show that $\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{-i\frac{\omega}{2}t}\\ e^{i\frac{\omega}{2}t} \end{pmatrix}$ is a solution to the Schrodinger equation.
b) Calculate $<\hat{S}_{x}>$ as a function of time with respect to this state.

We are told
$\hat{S}_{z} = \frac{\bar{h}}{2}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix},$$\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$

## Homework Equations

det($\hat{S}_{z} - λI)=0$

## The Attempt at a Solution

This was a previous exam example - we went over it in class but I got a little bit lost. I know it has to do with eigenvalues and eigenvectors, and I can get up to a certain point but then I get stuck, and I've no clue about where to start for b, we didn't get time to do that in class.

I get that det($\hat{S}_{z} - λI)=0$, and I know that ($\hat{S}_{z} - λI) = \begin{pmatrix} \frac{\bar{h}}{2}-λ&0\\ 0&-\frac{\bar{h}}{2}-λ \end{pmatrix}.$ Substituting this into det($\hat{S}_{z} - λI)=0$ gives $\frac{-\bar{h}^{2}}{4}+λ^{2}=0.$ Solving for λ gives $λ=\frac{\bar{h}}{2}$ or $λ=-\frac{\bar{h}}{2}$. Then I substitute this back in, so that if $λ=\frac{\bar{h}}{2}$ then $\begin{pmatrix} 0&0\\ 0&-\bar{h} \end{pmatrix} Z=0$ where Z is some vector. Also if $λ=-\frac{\bar{h}}{2}$ then $\begin{pmatrix} \bar{h}&0\\ 0&0 \end{pmatrix} Z=0$ where Z is some vector. This is where I get stuck - I can't seem to solve for vector Z, which takes the form of \begin{pmatrix}a\\b\end{pmatrix}, without just ending up with zeros. Not sure how to find a meaningful value of Z. And even if I had one, not sure how to bring that around to prove $\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{-i\frac{\omega}{2}t}\\ e^{i\frac{\omega}{2}t} \end{pmatrix}$ is a solution to the Schrodinger equation.

## The Attempt at a Solution

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fzero
Homework Helper
Gold Member
Then I substitute this back in, so that if $λ=\frac{\bar{h}}{2}$ then $\begin{pmatrix} 0&0\\ 0&-\bar{h} \end{pmatrix} Z=0$ where Z is some vector. Also if $λ=-\frac{\bar{h}}{2}$ then $\begin{pmatrix} \bar{h}&0\\ 0&0 \end{pmatrix} Z=0$ where Z is some vector. This is where I get stuck - I can't seem to solve for vector Z, which takes the form of \begin{pmatrix}a\\b\end{pmatrix}, without just ending up with zeros. Not sure how to find a meaningful value of Z.
Let's take one more step and write, for ##\lambda = \hbar/2##,

$$\begin{pmatrix} 0&0\\ 0&-\bar{h} \end{pmatrix} Z = \begin{pmatrix} 0&0\\ 0&-\bar{h} \end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} =\begin{pmatrix} 0&0\\ 0&-\bar{h} b \end{pmatrix} =0.$$

This equation implies that ##b=0##, but ##a## is not constrained. So an eigenvector of ##\hat{S}_z## with eigenvalue ##\lambda = + \hbar## is a multiple of

$$\chi_+ = \begin{pmatrix}1\\0\end{pmatrix}.$$

Similarly, an eigenvector with eigenvalue ##\lambda = + \hbar## is a multiple of (it'll help if you fill in the steps for yourself)

$$\chi_- = \begin{pmatrix}0\\1\end{pmatrix}.$$

These unit vectors are probably familiar to you from the discussion in class. Eigenvectors of ##\hat{H}## are also multiples of these, but the corresponding eigenvalue also involves a factor of ##\omega##.

And even if I had one, not sure how to bring that around to prove $\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{-i\frac{\omega}{2}t}\\ e^{i\frac{\omega}{2}t} \end{pmatrix}$ is a solution to the Schrodinger equation.
The time-dependent Schrodinger equation (TDSE) is

$$\hat{H} \psi = i\hbar \frac{\partial \psi}{\partial t}.$$

The idea is to write ##\psi ## as a linear combination of the eigenvectors of ##\hat{H}## that we found above:

$$\psi(t) = a(t) \chi_+ + b(t) \chi_- = \begin{pmatrix} a(t)\\ b(t) \end{pmatrix}.$$

When ##\hat{H}## acts on ##\chi_\pm##, we can substitute in the appropriate eigenvalue. The TDSE will then become a pair of independent equations for the functions ##a(t),b(t)##. You should see if you can make progress on the problem from this and post back if you're still having trouble,

Okay so, from there I got to $\hat{H} = ω\hat{S}_{z}$ so then $ω\hat{S}_{z} \begin{pmatrix} a(t)\\ b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$ so then
$\begin{pmatrix} \frac{ω \bar{h}}{2}&0\\ 0&-\frac{ω \bar{h}}{2} \end{pmatrix} \begin{pmatrix} a(t)\\ b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$
From here I can multiply it out to $\begin{pmatrix} \frac{iω}{2}a(t)\\ -\frac{iω}{2}b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$ and so I can solve for $\psi = \begin{pmatrix} e^{\frac{iωt}{2}}\\ e^{-\frac{iωt}{2}} \end{pmatrix}$

This looks much closer to the final result, but I can't seem to work out a) why my signs appear to be around the wrong way and b) where the $\frac{1}{\sqrt{2}}$ comes into play. :S

fzero
Homework Helper
Gold Member
Okay so, from there I got to $\hat{H} = ω\hat{S}_{z}$ so then $ω\hat{S}_{z} \begin{pmatrix} a(t)\\ b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$ so then
$\begin{pmatrix} \frac{ω \bar{h}}{2}&0\\ 0&-\frac{ω \bar{h}}{2} \end{pmatrix} \begin{pmatrix} a(t)\\ b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$
From here I can multiply it out to $\begin{pmatrix} \frac{iω}{2}a(t)\\ -\frac{iω}{2}b(t) \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}$ and so I can solve for $\psi = \begin{pmatrix} e^{\frac{iωt}{2}}\\ e^{-\frac{iωt}{2}} \end{pmatrix}$
You've introduced an extra factor of ##i## in the next to last equation. It should be

$$\begin{pmatrix} \frac{ω}{2}a(t)\\ -\frac{ω}{2}b(t) \end{pmatrix} = i\hbar\frac{d\psi}{dt}$$

This looks much closer to the final result, but I can't seem to work out a) why my signs appear to be around the wrong way and b) where the $\frac{1}{\sqrt{2}}$ comes into play. :S
The $\frac{1}{\sqrt{2}}$ was introduced to normalize the wavefunction.

Ah ok, the normalisation makes sense. Forgot about that. Also makes sense about the extra i - I stuffed up typing in the equation, it should have read:
$\begin{pmatrix} \frac{iω}{2}a(t)\\ -\frac{iω}{2}b(t)\\ \end{pmatrix} = \frac{\delta\psi}{\delta t}$
Still struggling to work out how I get to:
$\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{-\frac{iωt}{2}}\\ e^{\frac{iωt}{2}}\\ \end{pmatrix}$

I seem to keep arriving at:
$\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{\frac{iωt}{2}}\\ e^{-\frac{iωt}{2}}\\ \end{pmatrix}$ instead? Does the change in signs occur at some point in the normalisation?

fzero
Homework Helper
Gold Member
Ah ok, the normalisation makes sense. Forgot about that. Also makes sense about the extra i - I stuffed up typing in the equation, it should have read:
$\begin{pmatrix} \frac{iω}{2}a(t)\\ -\frac{iω}{2}b(t)\\ \end{pmatrix} = \frac{\delta\psi}{\delta t}$
Still struggling to work out how I get to:
$\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{-\frac{iωt}{2}}\\ e^{\frac{iωt}{2}}\\ \end{pmatrix}$

I seem to keep arriving at:
$\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{\frac{iωt}{2}}\\ e^{-\frac{iωt}{2}}\\ \end{pmatrix}$ instead? Does the change in signs occur at some point in the normalisation?
When you moved the ##i## from one side of the equation to the other, you missed the minus sign coming from ##1/i = - i##. You should have

$\begin{pmatrix} -\frac{iω}{2}a(t)\\ \frac{iω}{2}b(t)\\ \end{pmatrix} = \frac{\delta\psi}{\delta t}$

The first part all makes sense but now I'm stuck on part b) - calculate $<\hat{S}_{x}>$.
$\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$
So the expectation value = $\int^{\infty}_{-\infty} \psi*<\hat{S}_{x}>\psi dx$ so I get to $\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix} e^{i\frac{w}{2}t}\\ e^{-i\frac{w}{2}t} \end{pmatrix}\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} e^{-i\frac{w}{2}t}\\ e^{i\frac{w}{2}t} \end{pmatrix}$.
From here I get a bit stuck because if I multiply $<\hat{S}_{x}>$ by $\psi$ then I end up with $\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix} e^{i\frac{w}{2}t}\\ e^{-i\frac{w}{2}t} \end{pmatrix}\begin{pmatrix} e^{i\frac{w}{2}t}\\ e^{-i\frac{w}{2}t} \end{pmatrix}$ which comes out to
$\int^{\infty}_{-\infty}\frac{\bar{h}}{2}\psi*^{2}$. Is this the final answer? I feel like there should be a way to simplify, am I mixing up the calculations somewhere? If I can get to $\int^{\infty}_{-\infty} \psi\psi*$ then I know this cancels out to 1 since the solution is normalised.

fzero
Homework Helper
Gold Member
When the wavefunctions are spinors or vectors, instead of ##\psi^*## appearing in the inner product, we must have the adjoint wavefunction ##\psi^\dagger##. In addition to taking the complex conjugate, the adjoint also takes the transpose of the object: ##\psi^\dagger = (\psi^*)^T = (\psi^T)^*##. This is because the inner product ##\langle \psi | \chi \rangle## must be a c-number.

Ok, so I've followed this through and I end up with $\frac{1}{2}\frac{\bar{h}}{2}$$\int^{\infty}_{-\infty} \begin{pmatrix} e^{-i\frac{ω}{2}t}t&e^{i\frac{ω}{2}t} \end{pmatrix}\begin{pmatrix} e^{i\frac{ω}{2}t}\\ e^{-i\frac{ω}{2}t} \end{pmatrix}$ which then works out to be $\frac{\bar{h}}{2}$ ?

fzero
Homework Helper
Gold Member

$$\psi^* = \begin{pmatrix} e^{i\frac{w}{2}t}\\ e^{-i\frac{w}{2}t} \end{pmatrix},$$

which is correct. When we take the transpose of this, we have

$$\psi^\dagger = \begin{pmatrix} e^{i\frac{w}{2}t} & e^{-i\frac{w}{2}t} \end{pmatrix} .$$

When you put this back in the expectation value, you should write the result in terms of trig functions to verify that the expectation value is real.

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