Tag Archives: energy-time uncertainty principle

Uncertainty principle: a couple of examples from astronomy

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problems 5.14 – 5.15.

The uncertainty principle relates the standard deviations of two observables {A} and {B} to the expectation value of their commutator:

\displaystyle \sigma_{A}^{2}\sigma_{B}^{2}\ge\left(\frac{1}{2i}\left\langle \left[A,B\right]\right\rangle \right)^{2} \ \ \ \ \ (1)

For position {x} and momentum {p}, {\left[x,p\right]=i\hbar} so

\displaystyle \sigma_{x}\sigma_{p}\ge\frac{\hbar}{2} \ \ \ \ \ (2)

Example 1 In white dwarf stars, atoms become crushed together so that electrons and protons are much closer to each other than in ordinary hydrogen gas, where the mean radius of the electron’s orbit is the Bohr radius of {a=5.29\times10^{-11}\mbox{ m}}. In a white dwarf, this distance gets compressed to around {\sigma_{x}\approx1.5\times10^{-12}\mbox{ m}}. With the electron’s location thus localized, its momentum must be uncertain by an amount

\displaystyle \sigma_{p}\approx\frac{\hbar}{2\times1.5\times10^{-12}}=3.5\times10^{-23}\mbox{ kg m s}^{-1} \ \ \ \ \ (3)

The minimum average momentum must be equal to {\sigma_{p}} (otherwise the range of values for {p} would include negative values) so the electron’s minimum speed (assuming non-relativistic speeds) is

\displaystyle v_{min}=\frac{\sigma_{p}}{m_{e}}=\frac{3.5\times10^{-23}}{9.10938291\times10^{-31}}=3.86\times10^{7}\mbox{ m s}^{-1} \ \ \ \ \ (4)

This is about {0.13c} so we should probably use relativity to calculate a better value, but at least it gives an idea of how fast electrons must be moving in a white dwarf.

The energy-time uncertainty relation is a bit more subtle, since time in non-relativistic quantum mechanics is not an observable property of a quantum state; rather it’s a background parameter on which the quantum state depends. The energy-time uncertainty relation is usually given as

\displaystyle \Delta E\Delta t\ge\frac{\hbar}{2} \ \ \ \ \ (5)

where

\displaystyle \Delta t\equiv\frac{\sigma_{Q}}{\left|d\left\langle Q\right\rangle /dt\right|} \ \ \ \ \ (6)

with {Q} representing some arbitrary observable on the system. That is, to first order, {\Delta t} is the time interval during which {\left\langle Q\right\rangle } changes by one standard deviation. For a time-independent hamiltonian and a given set of initial conditions, the probabilities of finding the system in any given energy state do not depend on time, so {\Delta E} is constant, and serves as a constraint on the time scale over which other observables {Q} can change.

Conversely, if we can measure {\Delta t} for some observable (that is, if we can measure how fast some parameter of the system changes), we can get an estimate of {\Delta E}.

Example 2 An electron in the first excited state decays to the ground state in a time interval of around {10^{-8}\mbox{ s}}, by emitting a photon. Since such a decay is a change in an observable property of the system, we can use this time as an estimate of {\Delta t} and use it to derive an estimate of {\Delta E}, the standard deviation of the excited state energy.

\displaystyle \Delta E\approx\frac{\hbar}{2\Delta t}=5.27\times10^{-27}\mbox{ J}=3.29\times10^{-8}\mbox{ eV} \ \ \ \ \ (7)

This gives rise to a spread of wavelengths for the emitted photon:

\displaystyle E \displaystyle = \displaystyle h\nu=\frac{hc}{\lambda}\ \ \ \ \ (8)
\displaystyle \left|\Delta E\right| \displaystyle = \displaystyle hc\frac{\Delta\lambda}{\lambda^{2}} \ \ \ \ \ (9)

The transition {2\rightarrow1} is the first spectral line in the Lyman series with a wavelength of {\lambda_{2\rightarrow1}=121.6\mbox{ nm}} so, using {hc=1240\mbox{ eV nm}} we have

\displaystyle \Delta\lambda=3.29\times10^{-8}\frac{\left(121.6\right)^{2}}{1240}=3.92\times10^{-7}\mbox{ nm} \ \ \ \ \ (10)

This natural broadening of spectral lines would seem to be negligible.

Incidentally, this shows the inadequacy of the Schrödinger equation for studying the dynamics of electron energy levels, since the excited states of an atom are all eigenstates of the hamiltonian and thus should be stable. The fact that spontaneous decay occurs illustrates the need for quantum electrodynamics.

Translations in space and time

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.39.

Since {\hat{p}=(\hbar/i)\partial/\partial x}, we can write the exponential of the momentum operator by using a Taylor expansion:

\displaystyle  e^{i\hat{p}x_{0}/\hbar}=1+x_{0}\frac{\partial}{\partial x}+\frac{x_{0}^{2}}{2!}\frac{\partial^{2}}{\partial x^{2}}+\ldots \ \ \ \ \ (1)

where {x_{0}} is a constant. If we apply this operator to a function, we get

\displaystyle  e^{i\hat{p}x_{0}/\hbar}f(x)=f(x)+x_{0}\frac{\partial f}{\partial x}+\frac{x_{0}^{2}}{2!}\frac{\partial^{2}f}{\partial x^{2}}+\ldots \ \ \ \ \ (2)

which is the Taylor expansion of {f(x+x_{0})} about the point {x}, provided that the derivatives are evaluated at that point. Since this exponential operator effectively shifts the function by a distance {x_{0}}, the operator {p/\hbar} is called the generator of translations in space.

Similarly, for a function {\Psi(x,t)} that satisfies the time-dependent Schrodinger equation, we have

\displaystyle  i\hbar\frac{\partial\Psi(x,t)}{\partial t}=\hat{H}\Psi(x,t) \ \ \ \ \ (3)

so

\displaystyle  e^{-i\hat{H}t_{0}/\hbar}=1+t_{0}\frac{\partial}{\partial t}+\frac{t_{0}^{2}}{2!}\frac{\partial^{2}}{\partial t^{2}}+\ldots \ \ \ \ \ (4)

and

\displaystyle   e^{-i\hat{H}t_{0}/\hbar}\Psi(x,t) \displaystyle  = \displaystyle  \Psi(x,t)+t_{0}\frac{\partial\Psi}{\partial t}+\frac{t_{0}^{2}}{2!}\frac{\partial^{2}\Psi}{\partial t^{2}}+\ldots\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \Psi(x,t+t_{0}) \ \ \ \ \ (6)

which is the Taylor expansion with respect to time. The operator {-H/\hbar} is therefore known as the generator of translations in time.

The mean of an observable at time {t+t_{0}} is

\displaystyle   \langle Q\rangle_{t+t_{0}} \displaystyle  = \displaystyle  \langle\Psi(t+t_{0})|Q(t+t_{0})|\Psi(t+t_{0})\rangle\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \langle e^{-i\hat{H}t_{0}/\hbar}\Psi(t)|Q(t+t_{0})|e^{-i\hat{H}t_{0}/\hbar}\Psi(t)\rangle\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \langle\Psi(t)|e^{i\hat{H}t_{0}/\hbar}Q(t+t_{0})e^{-i\hat{H}t_{0}/\hbar}|\Psi(t)\rangle \ \ \ \ \ (9)

Expanding the two exponentials and the operator to first order in {t_{0}}, we get

\displaystyle   \langle Q\rangle_{t+t_{0}} \displaystyle  = \displaystyle  \left\langle \Psi(t)|\left(1+\frac{i}{\hbar}t_{0}\hat{H}\right)\left(Q(t)+t_{0}\frac{\partial Q}{\partial t}\right)\left(1-\frac{i}{\hbar}t_{0}\hat{H}\right)|\Psi(t)\right\rangle \ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \left\langle \Psi(t)|Q(t)+t_{0}\frac{\partial Q}{\partial t}+\frac{i}{\hbar}t_{0}(\hat{H}Q(t)-Q(t)\hat{H})|\Psi(t)\right\rangle +\mathcal{O}(t_{0}^{2})\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \langle Q\rangle_{t}+\left\langle \frac{\partial Q}{\partial t}\right\rangle t_{0}+\frac{i}{\hbar}t_{0}\langle[\hat{H},Q]\rangle+\mathcal{O}(t_{0}^{2}) \ \ \ \ \ (12)

where {\mathcal{O}(t_{0}^{2})} represents terms of second and higher order in {t_{0}}. Moving {\langle Q\rangle_{t}} to the LHS, dividing through by {t_{0}} and taking the limit as {t_{0}\rightarrow0} gives us the relation we got while studying the energy-time uncertainty principle:

\displaystyle  \frac{d\langle Q\rangle}{dt}=\left\langle \frac{\partial Q}{\partial t}\right\rangle +\frac{i}{\hbar}\langle[\hat{H},Q]\rangle \ \ \ \ \ (13)

Energy-time uncertainty: an alternative definition

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.32.

The energy-time uncertainty relation is derived by calculating the standard deviation of the energy{\sigma_{H}}, in terms of the rate of change of another observable {Q}, and the expression comes out to

\displaystyle  \sigma_{H}\sigma_{Q}\ge\frac{\hbar}{2}\left|\frac{d}{dt}\left\langle Q\right\rangle \right| \ \ \ \ \ (1)

From here we define {\Delta E\equiv\sigma_{H}} and the uncertainty in time is

\displaystyle  \Delta t\equiv\frac{\sigma_{Q}}{\left|d\left\langle Q\right\rangle /dt\right|} \ \ \ \ \ (2)

This is roughly the amount of time it takes {Q} to change by one standard deviation.

A variation on the energy-time relation is to define {\Delta t\equiv\tau/\pi}, where {\tau} is the time it takes for a wave function to change into another wave function that is orthogonal to the original function.

To see how this definition works in practice, we can start with a wave function that is a combination of two stationary states (for some arbitrary potential; the actual potential doesn’t matter as we’ll see). That is:

\displaystyle  \Psi(x,0)=\frac{1}{\sqrt{2}}(\psi_{1}(x)+\psi_{2}(x)) \ \ \ \ \ (3)

The wave function at the later time {\tau} is then:

\displaystyle  \Psi(x,\tau)=\frac{1}{\sqrt{2}}(\psi_{1}(x)e^{-iE_{1}\tau/\hbar}+\psi_{2}(x)e^{-iE_{2}\tau/\hbar}) \ \ \ \ \ (4)

For these two functions to be orthogonal, we must have:

\displaystyle   \langle\Psi(x,\tau)|\Psi(x,0)\rangle \displaystyle  = \displaystyle  0\ \ \ \ \ (5)
\displaystyle  \frac{1}{2}(e^{iE_{1}\tau/\hbar}+e^{iE_{2}\tau/\hbar}) \displaystyle  = \displaystyle  0\ \ \ \ \ (6)
\displaystyle  (1+e^{i(E_{2}-E_{1})\tau/\hbar}) \displaystyle  = \displaystyle  0\ \ \ \ \ (7)
\displaystyle  (E_{2}-E_{1})\tau/\hbar \displaystyle  = \displaystyle  \pi\ \ \ \ \ (8)
\displaystyle  \frac{\tau}{\pi}(E_{2}-E_{1}) \displaystyle  = \displaystyle  \hbar \ \ \ \ \ (9)

where we used the orthonormality of {\psi_{1}} and {\psi_{2}} in getting line 2.

Taking {\tau/\pi=\Delta t} and {E_{2}-E_{1}=\Delta E} gives the condition

\displaystyle  \Delta E\Delta t=\hbar>\frac{\hbar}{2} \ \ \ \ \ (10)

This is consistent with the original time-energy uncertainty principle. (Actually, to work out {\sigma_{H}} rather than just {\Delta E}, we’d need to know the specific potential being used.)

Virial theorem

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 3, Post 31.

While examining the energy-time uncertainty relation, we derived the equation:

\displaystyle \frac{d}{dt}\left\langle Q\right\rangle =\frac{i}{\hbar}\left\langle \left[H,Q\right]\right\rangle +\left\langle \frac{\partial Q}{\partial t}\right\rangle \ \ \ \ \ (1)

 

If we take {Q=xp}, we need the commutator {[\hat{H},xp]}. This is a straightforward calculation using derivatives, and remembering that {p=(\hbar/i)\partial/\partial x}. We also use

\displaystyle \hat{H} \displaystyle = \displaystyle \frac{p^{2}}{2m}+V\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x) \ \ \ \ \ (3)

assuming that the potential is time-independent. To make using the derivatives easier (especially when using the product rule), it is best to apply the commutator to some arbitrary function f. The result is

\displaystyle \left[\hat{H},xp\right]f \displaystyle = \displaystyle \left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\left(x\frac{\hbar}{i}\frac{\partial}{\partial x}\right)f-\left(x\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)f\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle -\frac{\hbar^{3}}{2im}\left(x\frac{\partial^{3}f}{\partial x^{3}}+2\frac{\partial^{2}f}{\partial x^{2}}\right)+\frac{\hbar}{i}xV\frac{\partial f}{\partial x}+\frac{\hbar^{3}}{2im}\left(x\frac{\partial^{3}f}{\partial x^{3}}\right)-\frac{\hbar}{i}x\left(\frac{\partial V}{\partial x}f+V\frac{\partial f}{\partial x}\right)\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle -\frac{\hbar^{3}}{im}\frac{\partial^{2}f}{\partial x^{2}}-\frac{\hbar}{i}x\frac{\partial V}{\partial x}f \ \ \ \ \ (6)

from which we get

\displaystyle \frac{i}{\hbar}[\hat{H},xp]=-\frac{\hbar^{2}}{m}\frac{\partial^{2}}{\partial x^{2}}-x\frac{\partial V}{\partial x} \ \ \ \ \ (7)

The first term on the RHS is {2T}, so from 1, taking means of both sides gives the result:

\displaystyle \frac{d}{dt}\langle xp\rangle=2\langle T\rangle-\left\langle x\frac{dV}{dx}\right\rangle \ \ \ \ \ (8)

In a stationary state, {\langle p\rangle=0} and {d\langle x\rangle/dt=0} (since the particle has no net motion) so
the LHS is zero in this case and

\displaystyle 2\langle T\rangle=\left\langle x\frac{dV}{dx}\right\rangle \ \ \ \ \ (9)

This is known as the virial theorem. Actually, the
theorem is more common in statistical mechanics, where its form is

\displaystyle 2\left\langle T\right\rangle =-\sum_{k=1}^{N}\left\langle \mathbf{F}_{k}\cdot\mathbf{r}_{k}\right\rangle \ \ \ \ \ (10)

where {\mathbf{F}_{k}} is the force acting on particle {k}, located at position {\mathbf{r}_{k}}. For a conservative force, the force can be expressed as the negative gradient of a potential, which gives us the form we have derived here. The curious name ‘virial’ comes from the Latin word vis, which means ‘energy’ or ‘force’.

For the harmonic oscillator, {V=m\omega^{2}\left\langle x^{2}\right\rangle /2}, so

\displaystyle \left\langle x\frac{dV}{dx}\right\rangle =m\omega^{2}\left\langle x^{2}\right\rangle \ \ \ \ \ (11)

so {\langle T\rangle=m\omega^{2}\left\langle x^{2}\right\rangle /2=\langle V\rangle}. This agrees with the result we obtained by directly calculating the mean values earlier.

Energy-time uncertainty principle – example

Required math: calculus, vectors

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Section 3.6; Problem 3.20.

The generalized uncertainty principle for two operators is

\displaystyle  \sigma_{A}^{2}\sigma_{B}^{2}\ge\left(\frac{1}{2i}\left\langle [\hat{A},\hat{B}]\right\rangle \right)^{2} \ \ \ \ \ (1)

In the derivation of the energy-time uncertainty principle we found that for an operator {Q} satisfies the equation

\displaystyle  \frac{d}{dt}\left\langle Q\right\rangle =\frac{i}{\hbar}\left\langle \left[H,Q\right]\right\rangle +\left\langle \frac{\partial Q}{\partial t}\right\rangle \ \ \ \ \ (2)

If we set {Q=x} (the position operator) then we get (since this operator does not depend explicitly on time):

\displaystyle  \frac{d}{dt}\left\langle x\right\rangle =\frac{i}{\hbar}\left\langle \left[H,x\right]\right\rangle \ \ \ \ \ (3)

Using the generalized uncertainty principle from above, we have

\displaystyle   \sigma_{H}^{2}\sigma_{x}^{2} \displaystyle  \ge \displaystyle  \left(-\frac{\hbar}{2}\frac{d\left\langle x\right\rangle }{dt}\right)^{2}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{4m^{2}}\left(m\frac{d\left\langle x\right\rangle }{dt}\right)^{2}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{4m^{2}}\left\langle p\right\rangle ^{2} \ \ \ \ \ (6)

This is the same relation as we discovered earlier by considering only the generalized uncertainty principle on its own.

Energy-time uncertainty principle: Gaussian free particle

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.19.

As another example of the energy-time uncertainty relation, we can look again at the example of a travelling free particle with a Gaussian wave packet. We have already worked out most of what we need to test the uncertainty relation:

\displaystyle   \langle x\rangle \displaystyle  = \displaystyle  \frac{l\hbar t}{m}\ \ \ \ \ (1)
\displaystyle  \langle x^{2}\rangle \displaystyle  = \displaystyle  \frac{1+(2a\hbar t/m)^{2}+a(2\hbar lt/m)^{2}}{4a}\ \ \ \ \ (2)
\displaystyle  \langle p^{2}\rangle \displaystyle  = \displaystyle  \hbar^{2}(a+l^{2}) \ \ \ \ \ (3)

From this we get

\displaystyle   \left\langle H\right\rangle \displaystyle  = \displaystyle  \frac{\left\langle p^{2}\right\rangle }{2m}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}(a+l^{2})}{2m} \ \ \ \ \ (5)

We still need {\left\langle H^{2}\right\rangle }. From our previous calculations, we have the wave function:

\displaystyle  \Psi(x,t)=\left(\frac{2a}{\pi}\right)^{1/4}\frac{e^{\left(-ax^{2}+i(lx-\hbar l^{2}t/2m)\right)/(1+2i\hbar at/m)}}{\sqrt{1+2i\hbar at/m}} \ \ \ \ \ (6)

We can calculate {\left\langle H^{2}\right\rangle } by direct integration, using Maple:

\displaystyle   \left\langle H^{2}\right\rangle \displaystyle  = \displaystyle  \frac{\hbar^{4}}{4m^{2}}\int_{-\infty}^{\infty}\left|\frac{d^{2}\Psi(x,t)}{dx^{2}}\right|^{2}dx\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{4}}{4m^{2}}\left(3a^{2}+6al^{2}+l^{4}\right) \ \ \ \ \ (8)

As a check on this result, we can work out the units (always a good test to make sure you haven’t dropped a factor somewhere). From the original wave function, since exponents must be dimensionless, we know that {a} has dimensions {distance^{-2}} and {l} has {distance^{-1}}. Planck’s constant has dimensions of {energy\times time}, so the expression above has overall units of {energy^{4}\times time^{4}\times mass^{-2}\times distance^{-4}=energy^{2}}. (Recall kinetic energy is {mv^{2}/2}.) It’s also worth noting that {\left\langle H^{2}\right\rangle } is independent of time.

From here, we can get {\sigma_{H}^{2}}:

\displaystyle   \sigma_{H}^{2} \displaystyle  = \displaystyle  \left\langle H^{2}\right\rangle -\left\langle H\right\rangle ^{2}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right) \ \ \ \ \ (10)

We also have, from above

\displaystyle   \sigma_{x}^{2} \displaystyle  = \displaystyle  \left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}} \ \ \ \ \ (12)

We’re trying to show that {\sigma_{H}\sigma_{x}\ge\frac{\hbar}{2}\frac{d\left\langle x\right\rangle }{dt}=l\hbar^{2}/2m} from above. So we want

\displaystyle   \sigma_{H}^{2}\sigma_{x}^{2} \displaystyle  \ge \displaystyle  \frac{l^{2}\hbar^{4}}{4m^{2}}\ \ \ \ \ (13)
\displaystyle  \frac{a\hbar^{4}}{2m^{2}}\left(a+2l^{2}\right)\frac{m^{2}+4\left(at\hbar\right)^{2}}{4am^{2}} \displaystyle  \ge \displaystyle  \frac{l^{2}\hbar^{4}}{4m^{2}} \ \ \ \ \ (14)

The minimum of the LHS occurs at {t=0} so if the inequality is true there, it is true always. In this case, it reduces to

\displaystyle   a+2l^{2} \displaystyle  \ge \displaystyle  2l^{2}\ \ \ \ \ (15)
\displaystyle  a \displaystyle  \ge \displaystyle  0 \ \ \ \ \ (16)

This final condition is certainly true (it is required for the Gaussian wave form to converge at large {x}), so the uncertainty condition is verified.

Energy-time uncertainty principle: infinite square well

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.18.

As an example of the energy-time uncertainty relation, we can look again at the example of a particle in the infinite square well, which starts off in a combination of the two lowest states:

\displaystyle  \Psi(x,0)=A\left[\psi_{1}(x)+\psi_{2}(x)\right] \ \ \ \ \ (1)

We would like to verify by explicit calculation that

\displaystyle  \sigma_{H}\sigma_{x}\ge\frac{\hbar}{2}\frac{d\left\langle x\right\rangle }{dt} \ \ \ \ \ (2)

Referring back to our calculations on that example, we can get some of the needed quantities immediately:

\displaystyle   \frac{d\langle x\rangle}{dt} \displaystyle  = \displaystyle  \frac{8\hbar}{3ma}\sin(3\omega t)\ \ \ \ \ (3)
\displaystyle  \langle H\rangle \displaystyle  = \displaystyle  \frac{5\pi^{2}\hbar^{2}}{4ma^{2}}\ \ \ \ \ (4)
\displaystyle  \langle x\rangle \displaystyle  = \displaystyle  \frac{a}{2}-\frac{16a}{9\pi^{2}}\cos(3\omega t)\ \ \ \ \ (5)
\displaystyle  \Psi(x,t) \displaystyle  = \displaystyle  \frac{\sqrt{2}}{2}\psi_{1}(x)e^{-i\omega t}+\frac{\sqrt{2}}{2}\psi_{2}(x)e^{-4i\omega t}\ \ \ \ \ (6)
\displaystyle  E_{i} \displaystyle  = \displaystyle  \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}\ \ \ \ \ (7)
\displaystyle  \psi_{n}(x) \displaystyle  = \displaystyle  \sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (8)

We can calculate {\sigma_{H}^{2}} using these equations:

\displaystyle   (\hat{H}-\langle H\rangle)\Psi \displaystyle  = \displaystyle  \frac{\sqrt{2}}{2}[(E_{1}-\langle H\rangle])\psi_{1}e^{-i\omega t}+(E_{2}-\langle H\rangle)\psi_{2}e^{-4i\omega t}]\ \ \ \ \ (9)
\displaystyle  \sigma_{H}^{2} \displaystyle  = \displaystyle  \langle(\hat{H}-\langle H\rangle)\Psi|(\hat{H}-\langle H\rangle)\Psi\rangle\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}[(E_{1}-\langle H\rangle])^{2}+(E_{2}-\langle H\rangle)^{2}]\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{9}{16}\left(\frac{\pi^{2}\hbar^{2}}{ma^{2}}\right)^{2} \ \ \ \ \ (12)

Thus

\displaystyle  \sigma_{H}=\frac{3}{4}\frac{\pi^{2}\hbar^{2}}{ma^{2}} \ \ \ \ \ (13)

To get {\sigma_{x}}, we can use the fact that, for any quantity {A}, {\sigma_{A}^{2}=\langle A^{2}\rangle-\langle A\rangle^{2}}. We already know {\langle x\rangle}, so we can calculate {\langle x^{2}\rangle} using integration. We get:

\displaystyle   \langle x^{2}\rangle \displaystyle  = \displaystyle  \frac{1}{2}\int_{0}^{a}(x^{2}\psi_{1}^{2}(x)+x^{2}\psi_{2}^{2}(x)+2x^{2}\psi_{1}\psi_{2}\cos(3\omega t))dx\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \frac{a^{2}}{144\pi^{2}}(48\pi^{2}-45-256\cos(3\omega t)) \ \ \ \ \ (15)

where we did the integral using Maple. We now get {\sigma_{x}} (after simplifying):

\displaystyle   \sigma_{x}^{2} \displaystyle  = \displaystyle  \langle x^{2}\rangle-\langle x\rangle^{2}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{a^{2}}{1296\pi^{4}}(108\pi^{4}-405\pi^{2}-4096\cos^{2}(3\omega t))\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  a^{2}\left[\frac{1}{12}-\frac{5}{16\pi^{2}}-\left(\frac{16}{9\pi^{2}}\cos3\omega t\right)^{2}\right] \ \ \ \ \ (18)

If we define

\displaystyle  \beta^{2}\equiv108\pi^{4}-405\pi^{2}-4096\cos^{2}(3\omega t) \ \ \ \ \ (19)


then

\displaystyle  \sigma_{x}=\frac{a}{36\pi^{2}}\beta \ \ \ \ \ (20)

and

\displaystyle   \sigma_{H}\sigma_{x} \displaystyle  = \displaystyle  \frac{1}{48}\frac{\hbar^{2}}{ma}\beta \ \ \ \ \ (21)

We would like to show that this satisfies the inequality

\displaystyle  \frac{1}{48}\frac{\hbar^{2}}{ma}\beta\geq\frac{\hbar}{2}\frac{d\langle x\rangle}{dt}=\frac{4\hbar^{2}}{3ma}\sin(3\omega t) \ \ \ \ \ (22)

which will be true if {\beta\geq64\sin(3\omega t)}, or {\beta^{2}\geq4096\sin^{2}(3\omega t)}. Substituting for {\beta} from 19 we get

\displaystyle   108\pi^{4}-405\pi^{2}-4096\cos^{2}(3\omega t) \displaystyle  \stackrel{?}{\geq} \displaystyle  4096\sin^{2}(3\omega t)\ \ \ \ \ (23)
\displaystyle  108\pi^{4}-405\pi^{2} \displaystyle  \stackrel{?}{\geq} \displaystyle  4096\cos^{2}(3\omega t)+4096\sin^{2}(3\omega t)\ \ \ \ \ (24)
\displaystyle  108\pi^{4}-405\pi^{2} \displaystyle  \stackrel{?}{\geq} \displaystyle  4096 \ \ \ \ \ (25)

Putting in the numbers on the LHS leads to 6522.992 > 4096, which is true so the uncertainty principle is satisfied for this case.