# Peeter Joot's (OLD) Blog.

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## My submission for PHY356 (Quantum Mechanics I) Problem Set 3.

Posted by peeterjoot on November 30, 2010

# Problem 1.

## Statement

A particle of mass $m$ is free to move along the x-direction such that $V(X)=0$. The state of the system is represented by the wavefunction Eq. (4.74)

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k x} e^{- i \omega t} f(k)\end{aligned} \hspace{\stretch{1}}(1.1)

with $f(k)$ given by Eq. (4.59).

\begin{aligned}f(k) &= N e^{-\alpha k^2}\end{aligned} \hspace{\stretch{1}}(1.2)

Note that I’ve inserted a $1/\sqrt{2\pi}$ factor above that isn’t in the text, because otherwise $\psi(x,t)$ will not be unit normalized (assuming $f(k)$ is normalized in wavenumber space).

\begin{itemize}
\item
(a) What is the group velocity associated with this state?
\item
(b) What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(c) What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(d) Explain the physical meaning of the above results.
\end{itemize}

## Solution

### (a). group velocity.

To calculate the group velocity we need to know the dependence of $\omega$ on $k$.

Let’s step back and consider the time evolution action on $\psi(x,0)$. For the free particle case we have

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} = -\frac{\hbar^2}{2m} \partial_{xx}.\end{aligned} \hspace{\stretch{1}}(1.3)

Writing $N' = N/\sqrt{2\pi}$ we have

\begin{aligned}-\frac{i t}{\hbar} H \psi(x,0) &= \frac{i t \hbar }{2m} N' \int_{-\infty}^\infty dk (i k)^2 e^{i k x - \alpha k^2} \\ &= N' \int_{-\infty}^\infty dk \frac{-i t \hbar k^2}{2m} e^{i k x - \alpha k^2}\end{aligned}

Each successive application of $-iHt/\hbar$ will introduce another power of $-it\hbar k^2/2 m$, so once we sum all the terms of the exponential series $U(t) = e^{-iHt/\hbar}$ we have

\begin{aligned}\psi(x,t) =N' \int_{-\infty}^\infty dk \exp\left( \frac{-i t \hbar k^2}{2m} + i k x - \alpha k^2 \right).\end{aligned} \hspace{\stretch{1}}(1.4)

Comparing with 1.1 we find

\begin{aligned}\omega(k) = \frac{\hbar k^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.5)

This completes this section of the problem since we are now able to calculate the group velocity

\begin{aligned}v_g = \frac{\partial {\omega(k)}}{\partial {k}} = \frac{\hbar k}{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

## (b). What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

In order to evaluate the probability, it looks desirable to evaluate the wave function integral 1.4.
Writing $2 \beta = i/(\alpha + i t \hbar/2m )$, the exponent of that integral is

\begin{aligned}-k^2 \left( \alpha + \frac{i t \hbar }{2m} \right) + i k x&=-\left( \alpha + \frac{i t \hbar }{2m} \right) \left( k^2 - \frac{i k x }{\alpha + \frac{i t \hbar }{2m} } \right) \\ &=-\frac{i}{2\beta} \left( (k - x \beta )^2 - x^2 \beta^2 \right)\end{aligned}

The $x^2$ portion of the exponential

\begin{aligned}\frac{i x^2 \beta^2}{2\beta} = \frac{i x^2 \beta}{2} = - \frac{x^2 }{4 (\alpha + i t \hbar /2m)}\end{aligned}

then comes out of the integral. We can also make a change of variables $q = k - x \beta$ to evaluate the remainder of the Gaussian and are left with

\begin{aligned}\psi(x,t) =N' \sqrt{ \frac{\pi}{\alpha + i t \hbar/2m} } \exp\left( - \frac{x^2 }{4 (\alpha + i t \hbar /2m)} \right).\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that from 1.2 we can compute $N = (2 \alpha/\pi)^{1/4}$, which could be substituted back into 1.7 if desired.

Our probability density is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 &=\frac{1}{{2 \pi}} N^2 {\left\lvert{ \frac{\pi}{\alpha + i t \hbar/2m} }\right\rvert} \exp\left( - \frac{x^2}{4} \left( \frac{1}{{(\alpha + i t \hbar /2m)}} + \frac{1}{{(\alpha - i t \hbar /2m)}} \right) \right) \\ &=\frac{1}{{2 \pi}} \sqrt{\frac{2 \alpha}{\pi} } \frac{\pi}{\sqrt{\alpha^2 + (t \hbar/2m)^2 }} \exp\left( - \frac{x^2}{4} \frac{1}{{\alpha^2 + (t \hbar/2m)^2 }} \left( \alpha - i t \hbar /2m + \alpha + i t \hbar /2m \right)\right) \\ &=\end{aligned}

With a final regrouping of terms, this is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 =\sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right).\end{aligned} \hspace{\stretch{1}}(1.8)

As a sanity check we observe that this integrates to unity for all $t$ as desired. The probability that we find the particle at position $x > x_0$ is then

\begin{aligned}P_{x>x_0}(t) = \sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\int_{x=x_0}^\infty dx \exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right)\end{aligned} \hspace{\stretch{1}}(1.9)

The only simplification we can make is to rewrite this in terms of the complementary error function

\begin{aligned}\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty e^{-t^2} dt.\end{aligned} \hspace{\stretch{1}}(1.10)

Writing

\begin{aligned}\beta(t) = \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 },\end{aligned} \hspace{\stretch{1}}(1.11)

we have

\begin{aligned}P_{x>x_0}(t_0) = \frac{1}{{2}} \text{erfc} \left( \sqrt{\beta(t_0)/2} x_0 \right)\end{aligned} \hspace{\stretch{1}}(1.12)

Sanity checking this result, we note that since $\text{erfc}(0) = 1$ the probability for finding the particle in the $x>0$ range is $1/2$ as expected.

## (c). What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

This unit length probability is thus

\begin{aligned}P_{x>x_0+1/2}(t_0) - P_{x>x_0-1/2}(t_0) &=\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0+\frac{1}{{2}} \right) \right) -\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0-\frac{1}{{2}} \right) \right) \end{aligned} \hspace{\stretch{1}}(1.13)

## (d). Explain the physical meaning of the above results.

To get an idea what the group velocity means, observe that we can write our wavefunction 1.1 as

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k (x - v_g t)} f(k)\end{aligned} \hspace{\stretch{1}}(1.14)

We see that the phase coefficient of the Gaussian $f(k)$ “moves” at the rate of the group velocity $v_g$. Also recall that in the text it is noted that the time dependent term 1.11 can be expressed in terms of position and momentum uncertainties $(\Delta x)^2$, and $(\Delta p)^2 = \hbar^2 (\Delta k)^2$. That is

\begin{aligned}\frac{1}{{\beta(t)}} = (\Delta x)^2 + \frac{(\Delta p)^2}{m^2} t^2 \equiv (\Delta x(t))^2\end{aligned} \hspace{\stretch{1}}(1.15)

This makes it evident that the probability density flattens and spreads over time with the rate equal to the uncertainty of the group velocity $\Delta p/m = \Delta v_g$ (since $v_g = \hbar k/m$). It is interesting that something as simple as this phase change results in a physically measurable phenomena. We see that a direct result of this linear with time phase change, we are less able to find the particle localized around it’s original time $x = 0$ position as more time elapses.

# Problem 2.

## Statement

A particle with intrinsic angular momentum or spin $s=1/2$ is prepared in the spin-up with respect to the z-direction state ${\lvert {f} \rangle}={\lvert {z+} \rangle}$. Determine

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.16)

and

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.17)

and explain what these relations say about the system.

## Solution: Uncertainty of $S_z$ with respect to ${\lvert {z+} \rangle}$

Noting that $S_z {\lvert {f} \rangle} = S_z {\lvert {z+} \rangle} = \hbar/2 {\lvert {z+} \rangle}$ we have

\begin{aligned}{\langle {f} \rvert} S_z {\lvert {f} \rangle} = \frac{\hbar}{2} \end{aligned} \hspace{\stretch{1}}(2.18)

The average outcome for many measurements of the physical quantity associated with the operator $S_z$ when the system has been prepared in the state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$ is $\hbar/2$.

\begin{aligned}\Bigl(S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \Bigr) {\lvert {f} \rangle}&= \frac{\hbar}{2} {\lvert {f} \rangle} -\frac{\hbar}{2} {\lvert {f} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.19)

We could also compute this from the matrix representations, but it is slightly more work.

Operating once more with $S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ on the zero ket vector still gives us zero, so we have zero in the root for 2.16

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = 0\end{aligned} \hspace{\stretch{1}}(2.20)

What does 2.20 say about the state of the system? Given many measurements of the physical quantity associated with the operator $V = (S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1})^2$, where the initial state of the system is always ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, then the average of the measurements of the physical quantity associated with $V$ is zero. We can think of the operator $V^{1/2} = S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ as a representation of the observable, “how different is the measured result from the average ${\langle {f} \rvert} S_z {\lvert {f} \rangle}$”.

So, given a system prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, and performance of repeated measurements capable of only examining spin-up, we find that the system is never any different than its initial spin-up state. We have no uncertainty that we will measure any difference from spin-up on average, when the system is prepared in the spin-up state.

## Solution: Uncertainty of $S_x$ with respect to ${\lvert {z+} \rangle}$

For this second part of the problem, we note that we can write

\begin{aligned}{\lvert {f} \rangle} = {\lvert {z+} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ).\end{aligned} \hspace{\stretch{1}}(2.21)

So the expectation value of $S_x$ with respect to this state is

\begin{aligned}{\langle {f} \rvert} S_x {\lvert {f} \rangle}&=\frac{1}{{2}}( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) S_x ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) \\ &=\hbar ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( {\lvert {x+} \rangle} - {\lvert {x-} \rangle} ) \\ &=\hbar ( 1 + 0 + 0 -1 ) \\ &= 0\end{aligned}

After repeated preparation of the system in state ${\lvert {f} \rangle}$, the average measurement of the physical quantity associated with operator $S_x$ is zero. In terms of the eigenstates for that operator ${\lvert {x+} \rangle}$ and ${\lvert {x-} \rangle}$ we have equal probability of measuring either given this particular initial system state.

For the variance calculation, this reduces our problem to the calculation of ${\langle {f} \rvert} S_x^2 {\lvert {f} \rangle}$, which is

\begin{aligned}{\langle {f} \rvert} S_x^2 {\lvert {f} \rangle} &=\frac{1}{{2}} \left( \frac{\hbar}{2} \right)^2 ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( (+1)^2 {\lvert {x+} \rangle} + (-1)^2 {\lvert {x-} \rangle} ) \\ &=\left( \frac{\hbar}{2} \right)^2,\end{aligned}

so for 2.22 we have

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = \frac{\hbar}{2}\end{aligned} \hspace{\stretch{1}}(2.22)

The average of the absolute magnitude of the physical quantity associated with operator $S_x$ is found to be $\hbar/2$ when repeated measurements are performed given a system initially prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$. We saw that the average value for the measurement of that physical quantity itself was zero, showing that we have equal probabilities of measuring either $\pm \hbar/2$ for this experiment. A measurement that would show the system was in the x-direction spin-up or spin-down states would find that these states are equi-probable.

I lost one mark on the group velocity response. Instead of 3.23 he wanted

\begin{aligned}v_g = {\left. \frac{\partial {\omega(k)}}{\partial {k}} \right\vert}_{k = k_0}= \frac{\hbar k_0}{m} = 0\end{aligned} \hspace{\stretch{1}}(3.23)

since $f(k)$ peaks at $k=0$.

I’ll have to go back and think about that a bit, because I’m unsure of the last bits of the reasoning there.

I also lost 0.5 and 0.25 (twice) because I didn’t explicitly state that the probability that the particle is at $x_0$, a specific single point, is zero. I thought that was obvious and didn’t have to be stated, but it appears expressing this explicitly is what he was looking for.

Curiously, one thing that I didn’t loose marks on was, the wrong answer for the probability per unit length. What he was actually asking for was the following

\begin{aligned}\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 - \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x_0, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.24)

That’s a whole lot more sensible seeming quantity to calculate than what I did, but I don’t think that I can be faulted too much since the phrase was never used in the text nor in the lectures.