Schrodinger solution spin half particles

[crowlma]

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}<br /> e^{-i\frac{\omega}{2}t}\\<br /> e^{i\frac{\omega}{2}t}<br /> \end{pmatrix} is a solution to the Schrodinger equation.
b) Calculate &lt;\hat{S}_{x}&gt; as a function of time with respect to this state.

We are told
\hat{S}_{z} = \frac{\bar{h}}{2}\begin{pmatrix}<br /> 1&amp;0\\<br /> 0&amp;-1<br /> \end{pmatrix},\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}<br /> 0&amp;1\\<br /> 1&amp;0<br /> \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}<br /> \frac{\bar{h}}{2}-λ&amp;0\\<br /> 0&amp;-\frac{\bar{h}}{2}-λ<br /> \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}<br /> 0&amp;0\\<br /> 0&amp;-\bar{h}<br /> \end{pmatrix} Z=0 where Z is some vector. Also if λ=-\frac{\bar{h}}{2} then \begin{pmatrix}<br /> \bar{h}&amp;0\\<br /> 0&amp;0<br /> \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}<br /> e^{-i\frac{\omega}{2}t}\\<br /> e^{i\frac{\omega}{2}t}<br /> \end{pmatrix} is a solution to the Schrodinger equation.

 

[fzero]

Then I substitute this back in, so that if λ=\frac{\bar{h}}{2} then \begin{pmatrix}<br /> 0&amp;0\\<br /> 0&amp;-\bar{h}<br /> \end{pmatrix} Z=0 where Z is some vector. Also if λ=-\frac{\bar{h}}{2} then \begin{pmatrix}<br /> \bar{h}&amp;0\\<br /> 0&amp;0<br /> \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}<br /> e^{-i\frac{\omega}{2}t}\\<br /> e^{i\frac{\omega}{2}t}<br /> \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,

 

[crowlma]

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

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

 

[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}<br /> \frac{iω}{2}a(t)\\<br /> -\frac{iω}{2}b(t)\\<br /> \end{pmatrix} = \frac{\delta\psi}{\delta t}
Still struggling to work out how I get to:
\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}<br /> e^{-\frac{iωt}{2}}\\<br /> e^{\frac{iωt}{2}}\\<br /> \end{pmatrix}

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

 

[fzero]

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}<br /> \frac{iω}{2}a(t)\\<br /> -\frac{iω}{2}b(t)\\<br /> \end{pmatrix} = \frac{\delta\psi}{\delta t}
Still struggling to work out how I get to:
\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}<br /> e^{-\frac{iωt}{2}}\\<br /> e^{\frac{iωt}{2}}\\<br /> \end{pmatrix}

I seem to keep arriving at:
\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}<br /> e^{\frac{iωt}{2}}\\<br /> e^{-\frac{iωt}{2}}\\<br /> \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}<br /> -\frac{iω}{2}a(t)\\<br /> \frac{iω}{2}b(t)\\<br /> \end{pmatrix} = \frac{\delta\psi}{\delta t}

 

[crowlma]

The first part all makes sense but now I'm stuck on part b) - calculate &lt;\hat{S}_{x}&gt;.
\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}<br /> 0&amp;1\\<br /> 1&amp;0<br /> \end{pmatrix}<br />
So the expectation value = \int^{\infty}_{-\infty} \psi*&lt;\hat{S}_{x}&gt;\psi dx so I get to \int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}<br /> e^{i\frac{w}{2}t}\\<br /> e^{-i\frac{w}{2}t}<br /> \end{pmatrix}\begin{pmatrix}<br /> 0&amp;1\\<br /> 1&amp;0<br /> \end{pmatrix}\begin{pmatrix}<br /> e^{-i\frac{w}{2}t}\\<br /> e^{i\frac{w}{2}t}<br /> \end{pmatrix}.
From here I get a bit stuck because if I multiply &lt;\hat{S}_{x}&gt; by \psi then I end up with \int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}<br /> e^{i\frac{w}{2}t}\\<br /> e^{-i\frac{w}{2}t}<br /> \end{pmatrix}\begin{pmatrix}<br /> e^{i\frac{w}{2}t}\\<br /> e^{-i\frac{w}{2}t}<br /> \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]

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.

 

[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}<br /> e^{-i\frac{ω}{2}t}t&amp;e^{i\frac{ω}{2}t}<br /> \end{pmatrix}\begin{pmatrix}<br /> e^{i\frac{ω}{2}t}\\<br /> e^{-i\frac{ω}{2}t}<br /> \end{pmatrix} which then works out to be \frac{\bar{h}}{2} ?

 

[fzero]

In post #7, you had

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

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.