Schrodinger solution spin half particles

1. Nov 4, 2013

crowlma

1. The problem statement, all variables and given/known data
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}$

2. Relevant equations
det($\hat{S}_{z} - λI)=0$

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 4, 2013

fzero

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 $\omega$.

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,

3. Nov 4, 2013

crowlma

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

4. Nov 4, 2013

fzero

You've introduced an extra factor of $i$ in the next to last equation. It should be

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

The $\frac{1}{\sqrt{2}}$ was introduced to normalize the wavefunction.

5. Nov 4, 2013

crowlma

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?

6. Nov 4, 2013

fzero

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

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

7. Nov 6, 2013

crowlma

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.

8. Nov 6, 2013

fzero

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.

9. Nov 6, 2013

crowlma

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 $\frac{\bar{h}}{2}$ ?

10. Nov 6, 2013

fzero

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