Homework due March 29 — MTH 496

Homework due March 29.

A PDF version of these exercises is available here.

The purpose of these exercises is to consider the free quantum evolution generated by the pure kinetic Hamiltonian:

\displaystyle  H=\frac{1}{2}P^{2}.

You may assume throughout that {\hbar=1}. Since {H\psi(x)=-\frac{1}{2}\frac{\partial^{2}\psi(x)}{\partial x^{2}}}, the Schroedinger equation for the evolution of the wave function is

\displaystyle  i\frac{\partial\psi_{t}(x)}{\partial t}=-\frac{1}{2}\frac{\partial^{2}\psi_{t}(x)}{\partial x^{2}}. \ \ \ \ \ (1)

Our goal is to solve (1) to find {\psi_{t}} in terms of the initial value of the wave function {\psi_{0}}.

Given a wave function {\psi(x)} in the position representation, denote its momentum representation by {\widetilde{\psi}\left(p\right)}. Recall that

\displaystyle  \widetilde{\psi}(p)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)e^{-ixp}dx.

In the momentum representation the Schroedinger equation reads

\displaystyle  i\frac{\partial\widetilde{\psi_{t}}\left(p\right)}{\partial t}=\frac{1}{2}p^{2}\widetilde{\psi_{t}}\left(p\right). \ \ \ \ \ (2)

Exercise 1 Show that the unique solution to (2) with given initial value {\widetilde{\psi}_{0}} is

\displaystyle  \widetilde{\psi}_{t}\left(p\right)=e^{-i\frac{1}{2}p^{2}t}\widetilde{\psi}_{0}\left(p\right). \ \ \ \ \ (3)

In the next few exercises, you will consider the evolution of certain specific initial wave functions. We start by defining these special functions — called coherent states — in the momentum representation

\displaystyle  \widetilde{\phi}^{x_{0},p_{0},z,\alpha}\left(p\right)=\left(\frac{\mathrm{Re}\ z}{\pi}\right)^{\frac{1}{4}}e^{-\frac{z}{2}\left(p-p_{0}\right)^{2}}e^{-ix_{0}p-i\alpha} \ \ \ \ \ (4)

where {x_{0},p_{0},\alpha\in\mathbb{R}} and {\mathrm{Re}\ z>0}. To express these wave functions in the position representation you will need the following formula (which you need not prove).

If {\mathrm{Re}\ z>0} and {w\in\mathbb{C}}, then

\displaystyle  \int_{-\infty}^{\infty}e^{-\frac{z}{2}\left(\lambda+w\right)^{2}}d\lambda=\frac{\sqrt{2\pi}}{\sqrt{z}}, \ \ \ \ \ (5)

where {\sqrt{z}} is the complex square root of {z} computed as follows, if {z=re^{i\theta}} with {-\frac{\pi}{2}<\theta<\frac{\pi}{2}} then

\displaystyle  \sqrt{z}=\sqrt{r}e^{i\frac{\theta}{2}}.

Exercise 2 Show that {\mathrm{Re}\ \frac{1}{z}>0} if {\mathrm{Re}\ z>0} and that the position space representation of {\widetilde{\phi}^{x_{0},p_{0},\sigma,\alpha}} is

\displaystyle  \phi^{x_{0},p_{0},z,\alpha}\left(x\right)=\left(\frac{\mathrm{Re}\ z}{\pi}\right)^{\frac{1}{4}}\frac{1}{\sqrt{z}}e^{-i\alpha}e^{-\frac{1}{2z}\left(x-x_{0}\right)^{2}}e^{ip_{0}\left(x-x_{0}\right)}. \ \ \ \ \ (6)

and that

\displaystyle  \int_{-\infty}^{\infty}\left|\phi^{x_{0},p_{0},z,\alpha}\left(x\right)\right|^{2}dx=1.

(Hint: expand and complete the square.)

Note that (6) can be expressed as saying {\phi^{x_{0},p_{0},z,\alpha}=\widetilde{\phi}^{-p_{0},x_{0},\frac{1}{z},\alpha'}} for suitable {\alpha'}. This wave function may be thought of as describing a particle with momentum approximately {p_{0}} and position approximately {x_{0}}. More specifically,

Exercise 3 Show that

\displaystyle  \left\langle P\phi^{x_{0},p_{0},z,\alpha},\phi^{x_{0},p_{0},z,\alpha}\right\rangle =p_{0},\quad\text{and}\quad\left\langle \left(P-p_{0}I\right)^{2}\phi^{x_{0},p_{0},z,\alpha},\phi^{x_{0},p_{0},z,\alpha}\right\rangle =\frac{1}{2\mathrm{Re}\ z},


\displaystyle  \left\langle Q\phi^{x_{0},p_{0},z,\alpha},\phi^{x_{0},p_{0},z,\alpha}\right\rangle =x_{0},\quad\text{and}\quad\left\langle \left(Q-x_{0}I\right)^{2}\phi^{x_{0},p_{0},z,\alpha},\phi^{x_{0},p_{0},z,\alpha}\right\rangle =\frac{\left|z\right|^{2}}{2\mathrm{Re}\ z}.

So the mean position and momentum in {\phi^{x_{0},p_{0},z,\alpha}} are {x_{0}} and {p_{0}} respectively, while the uncertainties are {\frac{\left|z\right|}{\sqrt{2\mathrm{Re}z}}} and {\frac{1}{\sqrt{2\mathrm{Re}z}}} respectively. Note that the product of the uncertainties is

\displaystyle  \Delta_{x_{0},p_{0},z,\alpha}P\Delta_{x_{0},p_{0},z,\alpha}Q=\frac{\left|z\right|}{2\left(\mathrm{Re}z\right)}=\frac{1}{2}\sqrt{1+\left(\frac{\mathrm{Im}\ z}{\mathrm{Re}\ z}\right)^{2}}\ge\frac{1}{2}

as required by the uncertainty principle.

Exercise 4 Now consider the Schroedinger equation (1) with initial condition {\psi_{0}=\phi^{x_{0},p_{0},z,\alpha}} for some {x_{0},p_{0},\alpha\in\mathbb{R}} and {z} with {\mathrm{Re}\ z>0}. Prove that the solution is given by

\displaystyle  \psi_{t}\left(x\right)=\phi^{x_{0}+tp_{0},p_{0},z+it,\alpha-\frac{1}{2}tp_{0}^{2}}(x).

Note that the mean position and mean momentum undergo the classical evolution {x=x_{0}+tp_{0}} and {p=p_{0}}. However the parameters {z} and {\alpha} also flow with the evolution. The change of {z} shows that the wave packet disperses, i.e., the uncertainty in {Q} diverges

\displaystyle  \Delta_{t}Q=\sqrt{\left\langle Q^{2}\psi_{t},\psi_{t}\right\rangle -\left\langle Q\psi_{t},\psi_{t}\right\rangle ^{2}}=\frac{\left|z+it\right|}{\sqrt{2\mathrm{Re}\ z}}\rightarrow\infty,\quad\text{as}\ t\rightarrow\infty.

By contrast, the uncertainty in {P} is constant in time.

In fact, position space dispersion of the wave function happens for any initial wave function. You will demonstrate it for initial wave functions {\psi_{0}} such that

\displaystyle  \sup_{x}\left(1+x^{2}\right)\left|\psi_{0}(x)\right|<\infty,\quad\text{and}\quad\sup_{p}(1+p^{2})\left|\widetilde{\psi_{0}}(p)\right|<\infty. \ \ \ \ \ (7)

Exercise 5 Assume (7). Prove that {\int_{-\infty}^{\infty}\left|\psi_{0}(x)\right|dx<\infty}, {\int_{-\infty}^{\infty}\left|\widetilde{\psi_{0}}(p)\right|<\infty}, {\int_{-\infty}^{\infty}\left|\psi\left(x\right)\right|^{2}dx<\infty}, and {\int_{-\infty}^{\infty}\left|\widetilde{\psi_{0}}(p)\right|^{2}d\lambda<\infty}.

Exercise 6 Assume (7) and show that

\displaystyle  \psi_{t}\left(x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\left(t\frac{p^{2}}{2}-xp\right)}\widetilde{\psi_{0}}\left(p\right)dp \ \ \ \ \ (8)

solves eq.\textasciitilde{}(1). (Hint: look at eq. (3).)

Exercise 7 Still assuming (7), argue that

\displaystyle  \psi_{t}(x)=\lim_{\delta\downarrow0}\int_{-\infty}^{\infty}U_{t,\delta}\left(x-y\right)\psi_{0}(y)dy


\displaystyle  U_{t,\delta}\left(x\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\left(t\frac{p^{2}}{2}-xp\right)-\frac{\delta}{2}p^{2}}dp

Exercise 8 Show that

\displaystyle  U_{t,\delta}\left(x\right)=\frac{1}{\sqrt{2\pi\sqrt{\delta^{2}+t^{2}}}}e^{-i\frac{\tan^{-1}\left(\frac{t}{\delta}\right)}{2}}e^{\frac{\left(it-\delta\right)x^{2}}{2\left(\delta^{2}+t^{2}\right)}}.

(Hint: complete the square and use (5).)

Exercise 9 Finally, assuming (7), prove that

\displaystyle  \psi_{t}\left(x\right)=\int_{-\infty}^{\infty}U_{t}\left(x-y\right)\psi_{0}\left(y\right) \ \ \ \ \ (9)

for {t\neq0}, where

\displaystyle  U_{t}\left(x\right)=\frac{e^{i\frac{x^{2}}{2t}}}{2\sqrt{\pi t}}\left(1-i\mathrm{sgn}t\right)

with {\mathrm{sgn}t=1} if {t>0} and {=-1} if {t<0}.

Exercise 10 Assuming (7), prove that

\displaystyle  \int_{-\infty}^{\infty}\left|\psi_{t}(x)\right|^{2}=\int_{-\infty}^{\infty}\left|\psi_{0}\left(x\right)\right|^{2} \ \ \ \ \ (10)

and that

\displaystyle  \sup_{x}\left|\psi_{t}\left(x\right)\right|\le\frac{1}{\sqrt{2\pi\left|t\right|}}\int_{-\infty}^{\infty}\left|\psi_{0}\left(x\right)\right|dx. \ \ \ \ \ (11)

(Hint: for the identity (10) use eq. (8) and facts about the Fourier transform; for the inequality (11) use eq. (9).)

You have assumed (7) in deriving eqs. (10) and (11). However, these expressions allow us to define a solution to the Schroedinger equation (1) for any integrable or square integrable {\psi_{0}}. Furthermore, (11) can be interpreted as dispersion of {\psi_{t}}, since the amplitude {\left|\psi_{t}\left(x\right)\right|} goes to zero as {t\rightarrow\infty}.

Here are two challenge problems, not required but if you want something even harder to think about then have a try:

  1. Prove that the uncertainty {\Delta_{t}Q\rightarrow\infty} as {t\rightarrow\infty} for any initial state {\psi_{0}}.
  2. Prove the magic formula (5) (actually this isn’t too hard if you know the right complex analysis).


2 thoughts on “Homework due March 29 — MTH 496

  1. Pingback: Correction to HW due March 22 — MTH 496 | Jeffrey Schenker

  2. Pingback: Corrections to HW and revised due date | Jeffrey Schenker

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