PHYS 481 - Quantum Mechanics

Assignment 02

Problem 1.7

Calculate \(d\left<p\right>/dt\). Answer:

\[\frac{d\left<p\right>}{dt} = \left<-\frac{\partial V}{\partial x}\right>\]

This is an instance of Ehrenfest’s theorem, which asserts that expectation values obey classical laws.

Response: Before proving this, we should derive a few quantities.

\[\begin{align}\frac{\partial \Psi}{\partial t} &= \frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi \\ \frac{\partial \Psi^*}{\partial t} &= -\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} + \frac{i}{\hbar}V\Psi\end{align}\]

\[\begin{align} \frac{d\left<p\right>}{dt} &= \frac{d}{dt}\left(-i\hbar\int_{-\infty}^{\infty}\Psi^*\frac{\partial\Psi}{\partial x} dx\right) \\ &= -i\hbar\int_{-\infty}^{\infty}\frac{d}{dt}\left(\Psi^*\frac{\partial\Psi}{\partial x}\right) dx \\ &= -i\hbar \int_{-\infty}^{\infty}\left(\Psi^*\frac{d}{dt}\frac{\partial\Psi}{\partial x} + \frac{\partial\Psi}{\partial x}\frac{\partial\Psi^*}{dt}\right) \\ &= -i\hbar\int_{-\infty}^{\infty}\left(\Psi^*\frac{\partial}{\partial x}\left(\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi\right) + \frac{\partial \Psi}{\partial x}\left(-\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} + \frac{i}{\hbar}V\Psi\right)\right) \\ &= -i\hbar\int_{-\infty}^{\infty}\left(\Psi^*\frac{i\hbar}{2m}\frac{\partial^3\Psi}{\partial x^3} - \frac{i}{\hbar}V\Psi^*\frac{\partial\Psi}{\partial x} - \frac{i}{\hbar}\frac{\partial V}{\partial x}\Psi^*\Psi - \frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}\frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V\Psi^*\frac{\partial\Psi}{\partial x}\right) \\ &= -i\hbar\left(\frac{-i\hbar}{2m}\int_{-\infty}^{\infty}\frac{\partial^2\Psi^*}{\partial x^2}\frac{\partial\Psi}{\partial x}dx + \frac{i\hbar}{2m}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^3\Psi}{\partial x^3}dx - \frac{i}{\hbar}\int_{-\infty}^{\infty}\frac{\partial V}{\partial x}\Psi^*\Psi dx\right) \\ &= -i\hbar \left(\frac{-i\hbar}{2m}\left(\frac{\partial \Psi}{\partial x}\frac{\partial \Psi^*}{\partial x}\right)\Big|_{-\infty}^{\infty} + \frac{i\hbar}{2m}\int_{-\infty}^{\infty}\frac{\partial \Psi^*}{\partial x}\frac{\partial^2\Psi}{\partial x^2}dx + \frac{i\hbar}{2m}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^3\Psi}{\partial x^3}dx - \frac{i}{\hbar}\int_{-\infty}^{\infty}\frac{\partial V}{\partial x}\Psi^*\Psi dx \right) \\ &= -i\hbar\left(\frac{i\hbar}{2m}\int_{-\infty}^{\infty}\frac{\partial \Psi^*}{\partial x}\frac{\partial^2\Psi}{\partial x^2}dx + \int_{-\infty}^{\infty}\left(\frac{i\hbar}{2m}\Psi^*\frac{\partial^3\Psi}{\partial x^3} - \frac{i}{\hbar}\frac{\partial V}{\partial x}\Psi^*\Psi\right)dx\right) \\ &= -i\hbar\left(\frac{i\hbar}{2m}\Psi^*\frac{\partial^2\Psi}{\partial x^2}\Big|_{-\infty}^{\infty} - \frac{i\hbar}{2m}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^3\Psi}{\partial x^3}dx + \frac{i\hbar}{2m}\int_{-\infty}^{\infty}\Psi^*\frac{\partial^3\Psi}{\partial x^3}dx - \frac{i}{\hbar}\int_{-\infty}^{\infty}\frac{\partial V}{\partial x}\Psi^*\Psi dx\right) \\ &= -\int_{-\infty}^{\infty}\Psi^*\frac{\partial V}{\partial x}\Psi dx \\ &= \left<-\frac{\partial V}{\partial x}\right>\end{align}\]

Problem 1.9

A particle of mass \(m\) has the wave function

\[\Psi(x,t) = Ae^{-a[(mx^2/\hbar) + it]},\]

where \(A\) and \(a\) are positive real constants.

1. Find \(A\).
2. For what potential energy function, \(V(x)\), is this a solution to the Schrodinger equation?
3. Calculate the expectation values of \(x\), \(x^2\), \(p\), and \(p^2\).
4. Find \(\sigma_x\) and \(\sigma_p\). Is their product consistent with the uncertainty principle?

Response: Normalizing \(\Psi\) to find \(A\) is done below.

\[\begin{align} 1 &= \int_{-\infty}^{\infty}\Psi^*\Psi dx \\ &= A^2 \int_{-\infty}^{\infty}e^{-2amx^2/\hbar} \\ &= A^2\left(\frac{\pi\hbar}{2ma}\right)^{1/2} \\ A &= \left(\frac{2ma}{\pi\hbar}\right)^{1/4}\end{align}\]

To find what potential energy function \(V(x)\) is a solution to the the Schrodinger equation, we need to find \(\frac{\partial \Psi}{\partial t}\) and \(\frac{\partial^2\Psi}{\partial x^2}\), then plug the results in to the Schrodinger equation, and solve for \(V(x)\), all of which are done below.

\[\begin{align} \frac{\partial \Psi}{\partial t} &= -ai\left(\frac{2ma}{\pi\hbar}\right)^{1/4}e^{-a[(mx^2/\hbar) + it]} \\ \frac{\partial^2\Psi}{\partial x^2} &= \left(\frac{2ma}{\pi\hbar}\right)^{1/4}\left(\frac{2max}{\hbar}\right)^2e^{-a[(mx^2/\hbar) + it]} + \left(\frac{2ma}{\pi\hbar}\right)^{1/4}\left(\frac{-2ma}{\hbar}\right)e^{-a[(mx^2/\hbar) + it]} \\ V(x) \Psi &= i\hbar\frac{\partial \Psi}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} \\ &= i\hbar\left(-ai\left(\frac{2ma}{\pi\hbar}\right)^{1/4}e^{-a[(mx^2/\hbar) + it]}\right) + \frac{\hbar^2}{2m}\left(\left(\frac{2ma}{\pi\hbar}\right)^{1/4}\left(\frac{2max}{\hbar}\right)^2e^{-a[(mx^2/\hbar) + it]} + \left(\frac{2ma}{\pi\hbar}\right)^{1/4}\left(\frac{-2ma}{\hbar}\right)e^{-a[(mx^2/\hbar) + it]}\right) \\ V(x) &= a\hbar\left(\frac{2ma}{\pi\hbar}\right)^{1/4}+ \frac{\hbar^2}{2m}\left(\frac{2ma}{\pi\hbar}\right)^{1/4}\left(\frac{2max}{\hbar}\right)^2 - \frac{\hbar^2}{2m}\left(\frac{2ma}{\hbar}\right)\left(\frac{2ma}{\hbar}\right) \\ V(x) &= 2ma^2x^2\end{align}\]

Calculating the expecation values of \(x\), \(x^2\), \(p\), and \(p^2\) is done below.

\[\begin{align} \left<x\right> &= \int_{-\infty}^{\infty}\Psi^*x\Psi dx \\ &= A^2\int_{-\infty}^{\infty}xe^{-2a(mx^2/\hbar)} \\ &= 0 \text{ because odd integrand over symmetric bounds} \\ \\ \left<x^2\right> &= \Psi^*x^2\Psi dx \\ &= \left(\frac{2am}{\pi\hbar}\right)^{1/2}\int_{-\infty}^{\infty}x^2e^{-2amx^2}dx \\ &= \left(\frac{2am}{\pi\hbar}\right)^{1/2}\left(\frac{\hbar}{4am}\right)\left(\frac{\pi\hbar}{2am}\right)^{1/2} \\ &= \frac{\hbar}{4am} \\ \\ \left<p\right> &= m\frac{d\left<x\right>}{dt} \\ &= 0 \text{ by Ehrenfest's Theorem} \\ \\ \left<p^2\right> &= \int_{-\infty}^{\infty}\Psi^*\hat{p}^2\Psi dx \\ &= -\hbar^2\int_{-\infty}^{\infty}\Psi^*\frac{\partial^2 \Psi}{\partial x^2}dx \\ &= \hbar^2\left(\frac{2am}{\pi\hbar}\right)^{1/2}\left(\frac{2am}{\hbar}\right)\int_{-\infty}^{\infty}\left(1 - \frac{2am}{\hbar}x^2\right)e^{-2amx^2/\hbar}dx\\ &= \hbar am\end{align}\]

It follows that \(\sigma_x\) and \(\sigma_p\) are

\[\begin{align}\sigma_x &= \sqrt{\left<x^2\right> - \left<x\right>^2} = \left(\frac{\hbar}{4am}\right)^{1/2} \\ \sigma_p &= \sqrt{\left<p^2\right> - \left<p\right>^2} = \left(\hbar a m\right)^{1/2} \end{align}\]

And their product is consistent with the uncertainty principle.

\[\sigma_x \sigma_p = \left(\frac{\hbar}{4am}\right)^{1/2} \left(\hbar a m\right)^{1/2} = \frac{\hbar}{2} \geq \frac{\hbar}{2}\]

Problem 1.14

Let \(P_{ab}(t)\) be the probability of finding the particle in the range \((a < x < b)\), at time \(t\).

1. Show that \[\frac{dP_{ab}}{dt} = J(a,t) - J(b, t),\] where \[J(x,t) \equiv \frac{i\hbar}{2m}\left(\Psi\frac{\partial \Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}\right).\] What are the units of \(J(x,t)\)? Comment: \(J\) is called the probability current because it tells you the rate at which probability is “flowing” past the point \(x\). If \(P_{ab}(t)\) is increasing, then more probability is flowing into the region at one end than flows out at the other.

2. Find the probability current for the wave function in Problem 1.9 (This is not a very pithy example, I’m afraid; we’ll encounter more substantial ones in due course.)

Response: For part a), we can derive the result the following way.

\[\begin{align} \frac{d}{dt}P_{ab} &= \frac{d}{dt}\int_{a}^{b}|\Psi|^2 dx \\ &= \int_a^b \frac{d}{dt}|\Psi|^2dx \\ &= \int_a^b \frac{\partial}{\partial x}\left(\frac{i\hbar}{2m}\left(\Psi^*\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x}\Psi\right)\right)dx \\ &= \int_a^b\frac{\partial}{\partial x}-J(x,t) dx \\ &= -J(x,t)\Big|_a^b \\ &= J(a,t) - J(b,t)\end{align}\]

Note that the units of \(J\) are \(\frac{1}{s}\).

For part b), if we call the real part of \(\Psi\) \(f(x)\) (to minimize the writing), we get the following probability current.

\[\begin{align}\Psi &= f(x)e^{-iat} \\ \Psi^* &= f(x)e^{iat} \\ \frac{\partial\Psi}{\partial x} &= f'(x)e^{-iat} \\ \frac{\partial\Psi^*}{\partial x} &= f'(x)e^{iat} \\ \\ J(x,t) &= \frac{i\hbar}{2m}\left(\Psi\frac{\partial \Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}\right) \\ &= \frac{i\hbar}{2m}\left(f(x)e^{-iat}f'(x)e^{iat} - f(x)e^{iat}f'(x)e^{-iat}\right) \\ &= 0\end{align}\]

Problem 1.16

A particle is represented (at time \(t=0\)) by the wavefunction \[\Psi(x,0) = \cases{A(a^2 - x^2), \hspace{1cm} -a\leq x\leq +a \\ 0, \hspace{3.3cm} \text{otherwise.}}\]

1. Determine the normalization constant \(A\).
2. What is the expectation value of \(x\)?
3. What is the expectation value of \(p\) (Note that you cannot get it from \(\left<p\right> = md\left<x\right>/dt\). Why not?)
4. Find the expectation value of \(x^2\).
5. Find the expectation value of \(p^2\).
6. Find the uncertainty in \(x\) (\(\sigma_x\)).
7. Find the uncertainty in \(p\) (\(\sigma_p\)).
8. Check that your results are consistent with the uncertainty principle.

Response: We can normalize \(\Psi\) to find \(A\) as follows:

\[\begin{align} 1 &= A^2\int_{-a}^{a}(a^2-x^2)^2dx \\ &= A^2\left(a^4\int_{-a}^{a}dx -2a^2\int_{-a}^{a}x^2dx + \int_{-a}^{a}x^4dx\right) \\ &= A^2\left(a^4x - 2a^2\frac{1}{3}x^3 + \frac{1}{3}x^5\right)\Big|_{-a}^{a} \\ &= A^2\left(2a^5-\frac{4}{3}a^5 + \frac{2}{3}a^5\right) \\ &= A^2\frac{16}{15}a^5 \\ A&= \frac{1}{4}\sqrt{\frac{15}{a^5}}\end{align}\]

From there we can find \(\left<x\right>\),

\[\begin{align}\left<x\right> &= \int_{-a}^{a}\Psi^*x\Psi dx \\ &= A^2\int_{-a}^{a}x(a^2-x^2)^2dx \\ &= 0 \text{ because odd integrand over symmetric bounds}\end{align}\]

\(\left<p\right>\) (which we can’t get from taking the time derivative because we are only given \(\Psi(t=0)\), not a distribution of \(\Psi\) across a range of \(t\)),

\[\begin{align}\left<p\right> &= \int_{-a}^{a}\Psi^*p\Psi dx \\ &= -i\hbar A^2\int_{-a}^{a}\underbrace{(a^2-x^2)}_{even}\underbrace{\frac{\partial}{\partial x}(a^2 - x^2)}_{odd}dx \\ &= 0 \text{ because odd integrand over symmetric bounds}\end{align}\]

\(\left<x^2\right>\),

\[\begin{align}\left<x^2\right> &= \int_{-a}^{a}\Psi^*x^2\Psi dx \\ &= A^2\left[a^4\int_{-a}^{a}x^2 dx - 2a^2\int_{-a}^{a}x^4 dx + \int_{-a}^{a} x^6 dx\right] \\ &= 2A^2\left[a^4\frac{1}{3}x^3 - \frac{2}{5}a^2x^5 + \frac{1}{7}a^7\right]_{0}^{a} \\ &= \frac{15}{8}a^2\left[\frac{1}{3} - \frac{2}{5} + \frac{1}{7}\right] \\ &= \frac{a^2}{7}\end{align}\]

and \(\left<p^2\right>\).

\[\begin{align}\left<p^2\right> &= \int_{-a}^{a}\Psi^*p^2\Psi dx \\ &= -\hbar^2 A^2\int_{-a}^{a}(a^2 - x^2)\frac{\partial^2}{\partial x^2}(a^2- x^2)dx \\ &= 2\hbar^2 A^2\int_{-a}^{a}(a^2-x^2)dx \\ &= \frac{15\hbar^2}{4a^5}\int_{0}^{a}(a^2-x^2)dx \\ &= \frac{15\hbar^2}{4a^5} \left(a^2 x - \frac{1}{3}x^3\right)_{0}^{a} \\ &= \frac{5\hbar^2}{2a^2}\end{align}\]

With those values, we can find \(\sigma_x\) and \(\sigma_p\), and compare them with the uncertainty principle.

\[\begin{align}\sigma_x &= \sqrt{\left<x^2\right> - \left<x\right>^2} = \frac{a}{\sqrt{7}} \\ \sigma_p &= \sqrt{\left<p^2\right> - \left<p\right>^2} = \sqrt{\frac{5}{2}}\frac{\hbar}{a} \\ \sigma_x\sigma_p &= \frac{a}{\sqrt{7}}\sqrt{\frac{5}{2}}\frac{\hbar}{a} = \sqrt{\frac{5}{14}}\hbar \geq \frac{\hbar}{2} \end{align}\]

Problem 1.17

Suppose you wanted to describe an unstable particle, that spontaneously disitegrates with a “lifetime” \(\tau\). In that case the total probability of finding the particle somewhere should not be constant, but should decrease at (say) an exponential rate:

\[P(t) \equiv \int_{-\infty}^{\infty}|\Psi(x,t)|^2dx = e^{-t/\tau}.\]

A crude way of achieving this result is as follows. In equation 1.24 we tacitly assumed that \(V\) (the potential energy) is real. That is certaintly reasonable, but it leads to the “conservation of probability” enshrined in Equations 1.27. What if we assign \(V\) to an imaginary part:

\[V = V_0 - i\Gamma,\]

where \(V_0\) is the true potential energy and \(\Gamma\) is a positive real constant?

1. Show that (in place of Equation 1.27) we now get \[\frac{dP}{dt} = -\frac{2\Gamma}{\hbar}P.\]
2. Solve for \(P(t)\), and find the lifetime of the particle in terms of \(\Gamma\).

Response: If we check out the “old” version of Equation 1.24

\[\frac{\partial\Psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + \frac{i}{\hbar}V\Psi^*\]

we can see that letting \(V\) be a complex function, then Equations 1.24 becomes the following

\[\frac{\partial\Psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + \frac{i}{\hbar}V^*\Psi^*\]

which causes 1.25, in which we apply the time derivative to \(|\Psi|^2\), to become the following

\[\begin{align} \frac{\partial}{\partial t}|\Psi|^2 &= \frac{i\hbar}{2m}\left(\Psi^*\frac{\partial^2\Psi}{\partial x^2} - \frac{\partial^2\Psi^*}{\partial x^2}\Psi\right) \\ &= \frac{\partial}{\partial x}\left[\frac{i\hbar}{2m}\left(\Psi^*\frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\right)\right] + \int\frac{i}{\hbar} |\Psi|^2 (V^* - V) dx \\ &= \frac{i}{\hbar}(V_0 + i\Gamma - V_0 + i\Gamma)\underbrace{\int |\Psi|^2 dx}_{P(t)} \\ \frac{dP}{dt} &= -\frac{2\Gamma}{\hbar}P\end{align}\]

Solving for \(P(t)\), we get the following relationship:

\[\begin{align} \int\frac{1}{P}dP &= \frac{-2\Gamma}{\hbar}\int dt \\ \ln P &= \frac{-2\Gamma t}{\hbar} + C \\ P(t) &= P(0)e^{-2\Gamma t/\hbar} \longrightarrow\tau \equiv\frac{\hbar}{2\Gamma}\end{align}\]

Problem 1.18

Very roughly speaking, quantum mechanics is relevant when the de Broglie wavelength of the particle in question (\(h/p\)) is greater than the characteristic size of the system (\(d\)). In thermal equilibrium at (Kelvin) temperature \(T\), the average kinetic energy of a particle is

\[\frac{p^2}{2m} = \frac{3}{2}k_B T\]

(where \(k_B\) is Boltzmann’s constant), so the typical de Broglie wavelength is

\[\lambda = \frac{h}{\sqrt{3mk_B T}}.\]

The purpose of this problem is to determine which systems will have to be treated quantum mechanically, and which can safely be described classically.

1. Solids. The lattice spacing in a typical solid is around \(d=0.3\) nm. Find the temperature below which the unbound electrons in a solid are quantum mechanical. Below what temperature are the nuclei in a solid quantum mechanical? (Use silicon as an example.) $$$$ Moral: the free electrons in a solid are always quantum mechanical; the nuclei are generally not quantum mechanical. The same goes for liquids (for which the interatomic spacing is roughly the same), with the exception of helium below 4 K.
2. Gases. For what temperatures are the atoms in an ideal gas at pressure \(P\) quantum mechanical? Hint: Use the ideal gas law (\(PV = Nk_B T\)) to deduce the interatomic spacing. $$$$ Answer: \(T<(1/k_B)(h^2/3m)^{3/5}P^{2/5}\). Obviously (for the gas to show quantum behavior) we want \(m\) to be as small as possible, and \(P\) as large as possible. Put in the numbers for helium at atmospheric pressure. Is hydrogen in outer space (where the interatomic spacing is about 1 cm and the temperature is about 3 K) quantum mechanical? (Assume it’s monatomic hydrogen, not H\(_2\).)

Response: If we rearrange the first equation given for \(p\), we can solve for the characteristic size of the system when things must be treated quantum mechanically, specifically

\[d < \frac{h}{\sqrt{3mk_B T}}\]

which corresponds to a temperature of

\[T < \frac{h^2}{3mk_Bd^2}\]

below which systems must be treated quantum mechanically.

Plugging in the mass of the electron (\(m_e = 9.11\cdot10^{-31}\) kg) and the mass of a silicon nucleus (\(m = 28.085m_p\), \(m_p = 1.67\cdot10^{-27}\) kg) at \(d = 0.3\) nm, we temperatures of \(T<1.3\cdot10^5\) K and \(T < 2.45\) K, respectively, which jives with the given moral that free electrons are always quantum mechanical and nuclei are generally not.

Looking into gasses, we can rewrite \(d\) in terms of the ideal gas law if we let \(N=1\) and \(V = d^3\).

\[PV = Nk_BT \longrightarrow d = \left(\frac{k_BT}{P}\right)^{1/3}\]

Plugging this result into our equation for \(T\), we get the following

\[\begin{align} T &< \frac{h^2}{3mk_Bd^2} = \frac{h^2}{3mk_B}\left(\frac{P}{k_B T}\right)^{2/5} \\ T^{5/3} &< \left(\frac{h^2P^{2/5}}{3mk_B^{5/3}}\right)^{5/3} \\ T &< \left(\frac{1}{k_B}\right)\left(\frac{h^2}{2m}\right)^{3/5}P^{2/5}\end{align}\]

Plugging in numbers for helium (\(m = 4m_p = 6.8\cdot10^{-27}\) kg) at atmospheric pressure (\(P=10^5\) N/m\(^2\)), using the equation we derived above, and hydrogen (\(m=1.67\cdot10^{-27}\) kg) in outerspace with \(d=0.01\) m, using \(T < \frac{h^2}{3mk_Bd^2}\), we get \(T<2.8\) K and \(T<6.35\cdot10^{-14}\) K for helium and hydrogen, respectively. These results imply that helium on earth are only quantum mechanical at very low temperatures, while hydrogen in space can be treated classically.