Solutions

Introduction to Quantum Mechanics

Chapter 2

Problem 2.1

Problem 2.2

Problem 2.3

Problem 2.4

Problem 2.5

Problem 2.1

Question (a)

The probability density:

|\Psi(x,t)|^2 = \bar \Psi \Psi = \bar \psi(x) e^{+\frac{iE}{\hbar}t} \psi(x) e^{-\frac{iE}{\hbar}t} = |\psi|^2

does not depend on t, as order to lead to normalizable solutions, E \in \mathbb R; considering as example an E = E_R + iE_I, the normalization condition is:

Considering t=0:

e^{0} \int_{-\infty}^\infty \bar \psi(x) \psi(x) \mathrm dx = 1

and therefore for t\neq 0:

e^{\frac{2E_I}{\hbar}t} = 1

This is possible only if E_I = 0 so E = E_R \in \mathbb R.

Question (b)

The time-independent wave function \psi(x) can always be taken to be real (whilst \Psi(x,t) is necessary complex); it does not means that every solution to the time-independent Schrödinger equation is real; if one isn’t, then it can be always expressed as a linear combination of solution with the same energy which are. Lets suppose \psi(x) is a solution of the equation for a particular E:

-\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi}{\partial x^2} + V\psi = E\psi

Considering \psi = a + ib and \bar \psi = a -ib, they both are solution of the above equation; considering the the following:

\begin{aligned} & \psi_1 \equiv \psi + \bar \psi = a + ib + a - ib = 2a = 2 \Re [\psi] \\ & \psi_2 \equiv i(\psi - \bar \psi) = i(a + ib - a + ib) = i2ib = -2 \Im [\psi] \end{aligned}

These are both real, plugging in the Schrödinger equation:

\begin{aligned} & -\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi_1}{\partial x^2} + V\psi_1 = E\psi_1 \\ & -\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi_2}{\partial x^2} + V\psi_2 = E\psi_2 \\ &-\frac{\hbar^2}{2m}\frac{\mathrm d^2 (\psi + \bar \psi)}{\partial x^2} + V(\psi + \bar \psi) = E(\psi + \bar \psi) \\ & -i\frac{\hbar^2}{2m}\frac{\mathrm d^2 (\psi - \bar \psi)}{\partial x^2} + iV(\psi - \bar \psi) = iE(\psi - \bar \psi) \end{aligned}

These equations are satisfied (as it can be split in the part for \psi and for \bar \psi), and the original complex \psi can be expressed as:

\psi_3 \equiv \frac{\psi_1 - i\psi_2}{2} = \frac{\psi + \bar \psi - i^2(\psi + \bar \psi)}{2} = \frac{2\psi}{2} = \psi

Question (c)

If V(x) is an even function (V(-x) = V(x)) then:

\begin{aligned} & -\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi(-x)}{\partial x^2} + V(-x)\psi(-x) = E\psi(-x) \\ & -\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi(y)}{\partial y^2} + V(y)\psi(y) = E\psi(y) \end{aligned}

The second equation is satisfied because V(x) is even by construction. Then taking the function

\begin{aligned} & \psi_+ \equiv \psi(x) + \psi(-x) \\ & \psi_- \equiv \psi(x) - \psi(-x) \end{aligned}

These are also a solution as they are linear combination of solutions; \psi_+ is even since:

\psi_+(-x) = \psi(-x) + \psi(-(-x)) = \psi(-x) + \psi(x)) = \psi_+(x)

and \psi_- is odd:

\psi_-(-x) = \psi(-x) - \psi(-(-x)) = \psi(-x) - \psi(x)) = -\psi_-(x)

and \psi(x) can be computed from these:

\psi(x) = \frac{\psi_+(x) + \psi_-(x)}{2} = \frac{\psi(x) + \psi(-x) + \psi(x) - \psi(-x)}{2} = \frac{2\psi(x)}{2}

And any solution can then be expressed as linear combination of even and odd functions.

Problem 2.2

The Energy E must exceed a minimum value V_{min} for every normalizable solution to the time-independent Schrödinger equation; writing it as:

\frac{\mathrm d^2 \psi}{\partial x^2} = \frac{2m}{\hbar^2}\left(V(x) - E\right)\psi

If E < V_{min} then:

\frac{\mathrm d^2 \psi}{\partial x^2} > 0; \quad x \in (-\infty, \infty)

since the second derivatives gives the curvature of a function, and since it is always positive it cannot go to zero at infinity and therefore cannot be normalized, as the integral cannot be equal to 1.

Problem 2.3

There are no acceptable solutions to the time-independent Schrödinger equation when E=0 or E < 0; this has been qualitatively proven with consideration of curvature of the function, but for the infinite well can be proven directly. In particular, as the expression for the energy is:

E = \frac{\hbar^2k^2}{2m} = \frac{\hbar^2n^2\pi^2}{2ma^2}

This expression must be real, and as the mass is positive and contains only quantity squared (real and positive quantities even for complex number) it cannot be negative.

It is also possible to prove it in a longer solving the Schrodinger equation and applying the boundary conditions; starting with E = 0, the equation becomes:

\frac{\mathrm d^2 \psi}{\partial x^2} = 0

This can be integrated twice with the solution:

\psi(x) = Ax + B

with A and B to be determined with the boundary conditions; setting them:

\begin{aligned} & \psi(0) = B = 0 \\ & \psi(a) = Aa = 0 \end{aligned}

So the solution is A = B = 0.

Considering the case E < 0, the equation becomes (keeping E positive and changing the sign):

\frac{\mathrm d^2 \psi}{\partial x^2} = \frac{2Em}{\hbar^2}\psi \equiv k^2\psi

The characteristic polynomial is:

\lambda^2 - k^2 = 0 \quad \Rightarrow \lambda = \pm k

and the solution is:

\psi(x) = Ae^{kx} + B e^{-kx} with A and B to be determined with the boundary conditions; setting them:

\begin{aligned} & \psi(0) = A + B = 0 \\ & \psi(a) = Ae^k + Be^{-k} = 0 \end{aligned}

Solving for A or B gives:

\begin{aligned} & A\left(e^{k} - e^{-k}\right)= 0 \\ & B\left(e^{-k} - e^{k}\right) = 0 \end{aligned}

This system has solution different from A = B =0 only if:

e^{-k} = e^{k} This equation has only the solution k =0, and k= \sqrt{\frac{2Em}{\hbar^2}} > 0 and therefore the solution is A = B = 0.

Problem 2.4

It is possible to compute \sigma_x based on the expectations \langle x\rangle and \langle x^2\rangle. The integrals used are computed at this link here.

Starting from \langle x\rangle:

\begin{aligned} \langle x \rangle = \int x |\Psi(x,t)|^2 \mathrm dx = & \int x |\psi(x)|^2 \mathrm dx = \int_{0}^a x \left[\sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right)\right]^2 \mathrm dx = \frac{2}{a} \int_{0}^a x \sin^2\left(\frac{n\pi}{a}x\right) \mathrm dx \\ = & \frac{2}{a}\left[\frac{x^2}{4} - \frac {a^2}{8n^2\pi^2}\cos\left(\frac{2n\pi}{a}x\right) - \frac {a}{4n\pi}x\sin\left(\frac{2n\pi}{a}x\right)\right]\bigg|_0^a \\ = & \frac{2}{a}\left[\frac{a^2}{4} - \frac {a^2}{8n^2\pi^2}\cos\left(2n\pi\right) + \frac {a^2}{8n^2\pi^2}\cos\left(0\right) - \frac {a}{4n\pi}a\sin\left(2n\pi\right) + \frac {a}{4n\pi}0\sin\left(0\right) \right] \\ = & \frac{2}{a}\left[\frac{a^2}{4} - \frac {a^2}{8n^2\pi^2} + \frac {a^2}{8n^2\pi^2} - 0 + 0 \right] = \frac{2}{a}\frac{a^2}{4} = \frac{a}{2} \end{aligned}

Then computing \langle x^2\rangle:

\begin{aligned} \langle x^2 \rangle = \int x^2 |\Psi(x,t)|^2 \mathrm dx = & \int x^2 |\psi(x)|^2 \mathrm dx = \int_{0}^a x^2 \left[\sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right)\right]^2 \mathrm dx = \frac{2}{a} \int_{0}^a x^2 \sin^2\left(\frac{n\pi}{a}x\right) \mathrm dx \\ = & \frac{2}{a}\left[\frac{x^3}{6} + \frac{a}{4n\pi} x^2 \sin \left(\frac{2n\pi}{a} x\right) - \frac {a^2}{4n^2\pi^2}x\cos \left(\frac{2n\pi}{a}x\right) -\frac{a^3}{8n^3\pi^3} \sin \left(\frac{2n\pi}{a}x\right) \right]\bigg|_0^a \\ = & \frac{2}{a}\left[ \frac{a^3}{6} - \frac{0}{6} + \frac{a}{4n\pi} a^2 \sin \left(2n\pi \right) - \frac{a}{4n\pi} 0 \sin \left(0\right) - \frac {a^2}{4n^2\pi^2}a\cos \left(2n\pi \right) \right.\\ & \left .+ \frac {a^2}{4n^2\pi^2} 0 \cos \left(0\right) -\frac{a^3}{8n^3\pi^3} \sin \left(2n\pi\right) + \frac{a^3}{8n^3\pi^3} \sin \left(0 \right) \right] \\ = & \frac{2}{a}\left[ \frac{a^3}{6} - 0 + 0 - 0 - \frac {a^2}{4n^2\pi^2}a\cos \left(2n\pi\right) + 0 - 0 + 0 \right] \\ = & \frac{a^2}{3} - \frac {a^2}{2n^2\pi^2} = a^2\left(\frac{1}{3} - \frac {1}{2n^2\pi^2}\right) \end{aligned}

Therefore the standard deviation \sigma_x is:

\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = \sqrt{a^2\left(\frac{1}{3} - \frac {1}{2n^2\pi^2}\right) - \frac{a^2}{4}} = a\sqrt{\frac{1}{12}-\frac {1}{2n^2\pi^2}}

Repeating the same for the momentum, starting from \langle p \rangle:

\begin{aligned} \langle p \rangle = & \int \bar \Psi(x,t) \left(-i\hbar\frac{\partial}{\partial x}\right)\Psi(x,t) \mathrm dx = \int \bar \psi(x) \left(-i\hbar\frac{\partial \psi(x)}{\partial x}\right) \mathrm dx \\ = & \frac{-2i\hbar n \pi}{a^2} \int_0^a \sin\left(\frac{n\pi}{a}x\right) \cos\left(\frac{n\pi}{a}x\right) \mathrm dx = \frac{-2i\hbar n \pi}{a^2} \left[ \frac{a}{2n\pi} \sin^2 \left(\frac{n\pi}{a}x\right) \right]\bigg|_0^a = 0 \end{aligned}

Then computing \langle p^2\rangle:

\begin{aligned} \langle p^2 \rangle = & \int \bar \Psi(x,t) \left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\Psi(x,t) \mathrm dx = \int \bar \psi(x) \left(-\hbar^2\frac{\partial^2 \psi(x)}{\partial x^2}\right) \mathrm dx \\ = & \frac{2\hbar^2 n^2 \pi^2}{a^3} \int_0^a \sin\left(\frac{n\pi}{a}x\right)^2 \mathrm dx = \frac{2\hbar^2 n^2 \pi^2}{a^3} \left[ \frac{x}{2} - \frac{a}{4n\pi} \sin\left(\frac{2n\pi}{a}x\right) \sin^2 \left(\frac{n\pi}{a}x\right) \right]\bigg|_0^a \\ = & \frac{2\hbar^2 n^2 \pi^2}{a^3}\left( \frac{a}{2} - 0 - 0 + 0 \right) = \left( \frac{\hbar n \pi}{a} \right)^2 \end{aligned}

Therefore the standard deviation \sigma_p is:

\sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} = \sqrt{\left( \frac{\hbar n \pi}{a} \right)^2} = \frac{\hbar n \pi}{a}

To check the uncertainty principle computing \sigma_x\sigma_p:

\sigma_x\sigma_p = a\sqrt{\frac{1}{12}-\frac {1}{2n^2\pi^2}} \times \frac{\hbar n \pi}{a} = \hbar n\pi\sqrt{\frac{1}{12}-\frac {1}{2n^2\pi^2}} = \hbar\sqrt{\frac{n^2\pi^2}{12}-\frac {1}{2}}

This number is smallest for n=1 and in this case it is:

\sigma_x\sigma_p = a\sqrt{\frac{1}{12}-\frac {1}{2n^2\pi^2}} \times \frac{\hbar n \pi}{a} = \hbar\sqrt{\frac{\pi^2}{12}-\frac{1}{2}} \approx \hbar \sqrt{0.32246 } \approx \hbar\; 0.56786

Since this quantity is greater than 0.5, the uncertainty principle is respected.

Problem 2.5

A particle in the infinite square well has as initial wave function a mixture of the first two states:

\Psi(x,0) = A\left(\psi_1(x) + \psi_2(x) \right)

Using the equation for the first two states explicitly it becomes:

\Psi(x,0) = A\left(\psi_1(x) + \psi_2(x) \right) = A\sqrt{\frac{2}{a}}\left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right]

The first step is to normalize this function:

\begin{aligned} \int_{-\infty}^\infty |\Psi(x,t)|^2 \mathrm dx & = \int_{-\infty}^\infty |\Psi(x,0)|^2 \mathrm dx = \int_0^a |A|^2 \left(\overline {\psi_1(x)} + \overline {\psi_2(x)} \right) \left(\psi_1(x) + \psi_2(x) \right) \mathrm dx \\ & = \int_0^a |A|^2 \frac{2}{a}\left\{\left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right]\left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right]\right\}\mathrm dx \\ & = |A|^2 \frac{2}{a} \int_0^a \left[ \sin^2\left(\frac{\pi}{a}x\right) + 2\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) \right]\mathrm dx \\ & = |A|^2 \frac{2}{a} \int_0^a \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) \right]\mathrm dx \\ & = |A|^2 \frac{2}{a} \left[ \frac{x}{2} - \frac{1}{4\pi}\sin\left(\frac{\pi}{a}x\right) + \frac{x}{2} - \frac{1}{8\pi}\sin\left(\frac{2\pi}{a}x\right) \right]\bigg|_0^a \\ & |A|^2 \frac{2}{a}a = |A|^2 \frac{1}{2} = 1 \end{aligned}

as \sin\left(\frac{\pi}{a}x\right) and \sin\left(\frac{2\pi}{a}x\right) are orthogonal and their product is 0; therefore:

A = \frac{1}{\sqrt 2}

and the wave function at t=0 is:

\Psi(x,0) = \frac{1}{\sqrt 2}\left(\psi_1(x) + \psi_2(x) \right) = \frac{1}{\sqrt a}\left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right]

The general solution is given by the infinite sum of:

\Psi(x,t) = \sum_{n=1}^\infty c_n \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right)e^{-i \frac{\hbar n^2\pi^2}{2ma^2}t}

The coefficient c_n can be computed using the Fourier transform; in the specific case as \Psi(x,0) is periodic, they only two non-null coefficients are known already:

c_1 = c_2 = \frac{1}{\sqrt 2}

As a verification they can be explicitly calculated; the generic formula is:

c_n = \int \overline{\psi_n(x)}f(x)\mathrm dx = \sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{n\pi}{a}x\right) \Psi(x,0) \mathrm dx

starting with c_1:

\begin{aligned} c_1 = & \sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{\pi}{a}x\right) \Psi(x,0) \mathrm dx = \frac{\sqrt 2}{a}\int_0^a \sin\left(\frac{\pi}{a}x\right) \left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right] \mathrm dx \\ = & \frac{\sqrt 2}{a}\int_0^a \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin\left(\frac{\pi}{a}x\right) \sin\left(\frac{2\pi}{a}x\right)\right] \mathrm dx = \frac{\sqrt 2}{a} \left\{ \left[ \frac{x}{2} - \frac{1}{4\pi}\sin\left(\frac{\pi}{a}x\right) \right]\bigg|_0^a + 0 \right\} \\ = & \frac{\sqrt 2}{a} \frac{a}{2} = \frac{\sqrt 2}{2} = \frac{1}{\sqrt 2} \end{aligned}

And similarly for c_2:

\begin{aligned} c_2 = & \sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{2\pi}{a}x\right) \Psi(x,0) \mathrm dx = \frac{\sqrt 2}{a}\int_0^a \sin\left(\frac{2\pi}{a}x\right) \left[ \sin\left(\frac{\pi}{a}x\right) + \sin\left(\frac{2\pi}{a}x\right)\right] \mathrm dx \\ = & \frac{\sqrt 2}{a}\int_0^a \left[ \sin\left(\frac{\pi}{a}x\right) \sin\left(\frac{2\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) \right] \mathrm dx = \frac{\sqrt 2}{a} \left\{ 0 + \left[ \frac{x}{2} - \frac{1}{8\pi}\sin\left(\frac{2\pi}{a}x\right) \right]\bigg|_0^a \right\} \\ = & \frac{\sqrt 2}{a} \frac{a}{2} = \frac{\sqrt 2}{2} = \frac{1}{\sqrt 2} \end{aligned}

Other coefficients are identically zero because of the orthogonality of the functions; the general solution it is then:

\Psi(x,t) = \frac{1}{\sqrt a} \left[ \sin\left(\frac{\pi}{a}x\right)e^{-i \frac{\hbar \pi^2}{2ma^2}t} + \sin\left(\frac{2\pi}{a}x\right)e^{-i \frac{\hbar 4\pi^2}{2ma^2}t} \right] = \frac{1}{\sqrt a} \left[ \sin\left(\frac{\pi}{a}x\right)e^{-i \omega t} + \sin\left(\frac{2\pi}{a}x\right)e^{-4i \omega t} \right]

where:

\omega \equiv \frac{\hbar \pi^2}{2ma^2}

Then |\Psi(x,t)|^2:

\begin{aligned} |\Psi(x,t)|^2 = & \overline {\Psi(x,t)} \Psi(x,t) =\frac{1}{a} \left\{\left[ \sin\left(\frac{\pi}{a}x\right)e^{i \omega t} + \sin\left(\frac{2\pi}{a}x\right)e^{4i \omega t} \right] \left[ \sin\left(\frac{\pi}{a}x\right)e^{-i \omega t} + \sin\left(\frac{2\pi}{a}x\right)e^{-4i \omega t} \right]\right\} \\ = & \frac{1}{a} \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin\left(\frac{\pi}{a}x\right)e^{i \omega t}\sin\left(\frac{2\pi}{a}x\right)e^{-4i \omega t} + \sin\left(\frac{2\pi}{a}x\right)e^{4i \omega t} \sin\left(\frac{\pi}{a}x\right)e^{-i \omega t} + \sin^2\left(\frac{2\pi}{a}x\right)\right] \\ = & \frac{1}{a} \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) + \sin\left(\frac{\pi}{a}x\right) \sin\left(\frac{2\pi}{a}x\right) \left(e^{3i \omega t} + e^{-3i \omega t}\right) \right]\\ = & \frac{1}{a} \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) + 2 \sin\left(\frac{\pi}{a}x\right) \sin\left(\frac{2\pi}{a}x\right) \cos(3\omega t) \right] \end{aligned}

As:

\begin{aligned} e^{3i \omega t} + e^{-3i \omega t} = & \cos(3 \omega t) + i\sin(3 \omega t) + \cos(-3 \omega t) + i\sin(-3 \omega t) \\ = & \cos(3 \omega t) + i\sin(3 \omega t) + \cos(3 \omega t) - i\sin(3 \omega t) = 2\cos(3 \omega t) \end{aligned}

Computing \langle x \rangle:

\begin{aligned} \langle x \rangle = & \int_{-\infty}^{\infty} x |\Psi(x,t)|^2 \mathrm dx \\ = & \frac{1}{a}\int_0^a x \left[ \sin^2\left(\frac{\pi}{a}x\right) + \sin^2\left(\frac{2\pi}{a}x\right) + 2 \sin\left(\frac{\pi}{a}x\right) \sin\left(\frac{2\pi}{a}x\right) \cos(3\omega t) \right] \mathrm dx \\ = & \frac{1}{a} \left\{\left[ \frac{x^2}{4} - \frac {a^2}{8\pi^2}\cos\left(\frac{2\pi}{a}x\right) - \frac {a}{4\pi}x\sin \left(\frac{2\pi}{a}x\right) \right]\bigg|_0^a \right. \\ & + \left[\frac{x^2}{4} - \frac {a^2}{16\pi^2}\cos\left(\frac{4\pi}{a}x\right) - \frac {a}{8\pi}x\sin \left(\frac{4\pi}{a}x\right) \right]\bigg|_0^a\\ & + 2\cos(3\omega t)\frac{1}{2}\left[ \frac {a}{\pi}x \sin \left(\frac{\pi}{a}x\right) - \frac {a}{3\pi}x\sin \left(\frac{3\pi}{a}x\right) \right. \\ & \left. \left. + \frac {a^2}{\pi^2}\cos\left(\frac{\pi}{a}x\right) - \frac {a^2}{9\pi^2}\cos\left(\frac{3\pi}{a}x\right) \right]\bigg|_0^a \right\} \\ = & \frac{1}{a} \left\{\left[ \frac{a^2}{4} - \frac{0}{4} - \frac {a^2}{8\pi^2}\cos\left(2\pi \right) + \frac {a^2}{8\pi^2}\cos\left(0\right) - \frac {a}{4\pi}x\sin \left(2\pi\right) + \frac {a}{4\pi}x\sin \left(0\right) \right] \right. \\ & + \left[\frac{a^2}{4} - \frac{0}{4} - \frac {a^2}{16\pi^2}\cos\left(4\pi \right) + \frac {a^2}{16\pi^2}\cos\left(0\right) - \frac {a}{8\pi}x\sin \left(4\pi\right) + \frac {a}{8\pi}x\sin \left(0\right) \right]\\ & + \cos(3\omega t)\left[ \frac {a}{\pi}a \sin \left(\pi\right) - \frac {a}{\pi}0 \sin \left(0\right) - \frac {a}{3\pi}a\sin \left(3\pi\right) + \frac {a}{3\pi}a\sin \left(0\right) \right. \\ & \left. \left. + \frac {a^2}{\pi^2}\cos\left(\pi\right) - \frac {a^2}{\pi^2}\cos\left(0\right) - \frac {a^2}{9\pi^2}\cos\left(3\pi \right) + \frac {a^2}{9\pi^2}\cos\left(0\right) \right] \right\} \\ = & \frac{1}{a} \left\{\left[ \frac{a^2}{4} - \frac {a^2}{8\pi^2} + \frac {a^2}{8\pi^2} \right] + \left[\frac{a^2}{4} - \frac {a^2}{16\pi^2} + \frac {a^2}{16\pi^2} \right] \right. \\ & \left. + \cos(3\omega t)\left[ -\frac {a^2}{\pi^2} - \frac {a^2}{\pi^2} + \frac {a^2}{9\pi^2} + \frac {a^2}{9\pi^2} \right] \right\} \\ = & \frac{1}{a}\left[\frac{a^2}{2}+\cos(3\omega t) \frac {2a^2}{\pi^2}\left(-1 + \frac{1}{9}\right)\right] = a\left[\frac{1}{2} - \frac{16}{9\pi^2} \cos(3\omega t)\right] \end{aligned}

So:

\langle x \rangle = a\left[\frac{1}{2} - \frac{16}{9\pi^2} \cos(3\omega t)\right] = a\left[\frac{1}{2} - \frac{16}{9\pi^2} \cos\left(3\frac{\hbar\pi^2}{2ma^2} t\right)\right]

The amplitude of the oscillation in the given expression is determined by the coefficient of the cosine term, which represents the maximum deviation from the mean position in an oscillatory motion; the amplitude is given by the coefficient of the \cos(3\omega t) term. This coefficient is \frac{16}{9\pi^2}. Thus, the amplitude of the oscillation is:

\frac{16}{9\pi^2}a \approx 0.180126 \; a < \frac{a}{2}

The angular frequency in an oscillatory motion is given by the coefficient of t in the argument of the cosine or sine function, in this case:

3\omega = 3\frac{\hbar\pi^2}{2ma^2}

To compute the average of the momentum \langle p \rangle it is not needed to compute directly, but it is possible to apply the formula:

\begin{aligned} \langle p \rangle = & m \langle v \rangle = m \frac{\mathrm d \langle x \rangle}{\mathrm dt} = ma \frac{\mathrm d}{\mathrm dt} \left[\frac{1}{2} - \frac{16}{9\pi^2} \cos(3\omega t)\right] = -\frac{16ma}{9\pi^2}(-3\omega\sin(3\omega t)) = \frac{16ma}{9\pi^2}3\frac{\hbar\pi^2}{2ma^2}\sin(3\omega t) \\ = & \frac{8\hbar}{3a}\sin(3\omega t) = \frac{8\hbar}{3a}\sin\left(3\frac{\hbar\pi^2}{2ma^2} t\right) \end{aligned}

Finally, computing the energies from the formula:

E = \frac{\hbar^2k^2}{2m} = \frac{\hbar^2n^2\pi^2}{2ma^2}

gives:

E_1 = \frac{\hbar^2\pi^2}{2ma} \quad E_2 = \frac{2\hbar^2\pi^2}{ma}

Their expectations is:

\langle E_1 \rangle = |c_1|^2 = \frac{1}{2} \quad \langle E_2 \rangle = |c_2|^2 = \frac{1}{2}

The total energy is:

\langle \mathbf H \rangle = \sum_{n=1}^\infty |c_n|^2 E_n = \sum_{n=1}^2 |c_n|^2 E_n = \frac{1}{2} \left(E_1 + E_2\right) = \frac{1}{2}\frac{\hbar^2\pi^2}{2ma} + \frac{1}{2}\frac{2\hbar^2\pi^2}{ma} = \frac{5\hbar^2\pi^2}{4ma}

The probability of getting the state E_1 or E_2 is 50\% and \langle \mathbf H \rangle is the average of these two states.

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