Consider the following wave functions, each describing a particle in one dimension and defined over all space (i.e., \(-\infty < x < \infty\)), not confined to an infinite square well.
\(\psi_a(x) = A e^{-x^2/3}\)
\(\psi_b(x) = B \frac{1}{x^2+2} \)
For each wave function:
For: \(\psi_a(x) = A e^{-x^2/3}\)
(a) Normalization: \begin{eqnarray*} \left\langle {\psi_a}\middle|{\psi_a}\right\rangle &=&1\\ &=& \int_{-\infty}^{\infty} \psi_a^*(x)\; \psi_a(x) dx\\ &=& \int_{-\infty}^{\infty} |A|^2 \left(e^{-x^2/3}\right)^2 dx\\ &=& \int_{-\infty}^{\infty} |A|^2 e^{-2x^2/3} dx\\ &=& |A|^2 \sqrt{\frac{3\pi}{2}}\\[6pt] A &=& \left(\frac{2}{3\pi}\right)^{1/4} \\ \end{eqnarray*} Therefore, \(\psi_a(x)=\left(\frac{2}{3\pi}\right)^{1/4} e^{-x^2/3}\)
(b) \begin{eqnarray*} \mbox{Probability} &=& \int_{0}^{1} \psi_a^*(x)\; \psi_a(x) dx\\ &=& \int_{0}^{1} \sqrt{\frac{2}{3\pi}} \left(e^{-x^2/3}\right)^2 dx\\ &=& \frac{1}{2}\mbox{Erf}\left(\sqrt{\frac{2}{3}}\right)\\ &\approx& 0.375 \end{eqnarray*}
For: \(\psi_b(x) = B \frac{1}{x^2+2} \)
(a) Normalization: \begin{eqnarray*} \left\langle {\psi_b}\middle|{\psi_b}\right\rangle &=&1\\ &=& \int_{-\infty}^{\infty} \psi_b^*(x)\; \psi_b(x) dx\\ &=& \int_{-\infty}^{\infty} |B|^2 \left(\frac{1}{x^2+2}\right)^2 dx\\ &=& \int_{-\infty}^{\infty} |B|^2 \frac{1}{(x^2+2)^2} dx\\ &=& |B|^2 \frac{\pi}{4\sqrt{2}}\\[6pt] B &=& \sqrt{\frac{4\sqrt{2}}{\pi}} \\ \end{eqnarray*} Therefore, \(\psi_b(x)=\sqrt{\frac{4\sqrt{2}}{\pi}} \frac{1}{x^2+2}\)(b) \begin{eqnarray*} \mbox{Probability} &=&\int_{0}^{1} \psi_b^*(x)\; \psi_b(x) dx\\ &=& \int_{0}^{1} {\frac{4\sqrt{2}}{\pi}} \left(\frac{1}{x^2+2}\right)^2 dx\\ &=& \frac{\sqrt{2}+3\arctan{(\frac{1}{\sqrt{2}})}}{3\pi}\\ &\approx& 0.35 \end{eqnarray*}
A particle in an infinite square well potential has an initial state vector \[\left|{\Psi(t=0)}\right\rangle = A\big(\left|{\phi_1}\right\rangle -\left|{\phi_2}\right\rangle +i\left|{\phi_3}\right\rangle \big)\]
where \(|\phi_1\rangle,|\phi_2\rangle\), and \(|\phi_3\rangle \) are the first three energy eigenstates (i.e., \(n=1,2,3\)).
Determine \(A\).
\begin{align*} 1 &= \left\langle {\Psi}\middle|{\Psi}\right\rangle \\ &= A^*\big(\left\langle {\phi_1}\right|-\left\langle {\phi_2}\right|-i\left\langle {\phi_3}\right|\big)A\big(\left|{\phi_1}\right\rangle -\left|{\phi_2}\right\rangle +i\left|{\phi_3}\right\rangle \big)\\ &= |A|^2 (3)\\ A &= \frac{1}{\sqrt{3}} \end{align*}
Write the initial state \(\left|{\Psi(t=0)}\right\rangle \) in wavefunction from.
To write the state in the position representation, use \[ \psi(x,0)=\langle x|\Psi(t=0)\rangle. \] Given \[ |\Psi(t=0)\rangle=A\big(|\phi_1\rangle-|\phi_2\rangle+i|\phi_3\rangle\big), \] we have \[ \psi(x,0)=A\big(\langle x|\phi_1\rangle-\langle x|\phi_2\rangle+i\langle x|\phi_3\rangle\big) = A\big(\phi_1(x)-\phi_2(x)+i\phi_3(x)\big). \] For the infinite square well on \(0<x<L\), \[ \phi_n(x)=\langle x|\phi_n\rangle=\sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right)\quad (0<x<L), \] and \(\phi_n(x)=0\) outside the well. Using \(A=\frac{1}{\sqrt{3}}\) from part (a), \[ \psi(x,0)=\frac{1}{\sqrt{3}}\sqrt{\frac{2}{L}} \left[\sin\!\left(\frac{\pi x}{L}\right)-\sin\!\left(\frac{2\pi x}{L}\right)+i\sin\!\left(\frac{3\pi x}{L}\right)\right], \quad 0<x<L, \] and \[ \psi(x,0)=0\quad \text{for } x\le 0 \text{ or } x\ge L. \]
At time \(t=0\), if an energy measurement is performed, what are the possible energy values, and with what probability would each possible value be obtained?
Since this is an infinite square well, energies are \(E_n = \frac{n^2\pi^2\hbar^2}{mL^2}\). Looking at the kets to determine \(n\) tells me the energy. Taking the norm square of the inner product to get the probabilities:
\begin{align*} \mathcal{P}\Big(E=\frac{\pi^2\hbar^2}{2mL^2}\Big) &= \Big| \left\langle {\phi_1}\middle|{\Psi}\right\rangle \Big|^2\\ &= \frac{1}{3}\\ \mathcal{P}\Big(E=\frac{4\pi^2\hbar^2}{2mL^2}\Big) &= \Big| \left\langle {\phi_2}\middle|{\Psi}\right\rangle \Big|^2\\ &= \frac{1}{3}\\ \mathcal{P}\Big(E=\frac{9\pi^2\hbar^2}{2mL^2}\Big) &= \Big| \left\langle {\phi_3}\middle|{\Psi}\right\rangle \Big|^2\\ &= \frac{1}{3}\\ \end{align*}
What is the expectation value of the energy of this particle at \(t=0\)?
The average value (expectation value) of the energy is:
\begin{align*} \langle H \rangle &= \sum_{n=1}^{3} \mathcal{P}(E_n)\;E_n\\ &= \frac{1}{3}\frac{\pi^2\hbar^2}{2mL^2} + \frac{1}{3}\frac{4\pi^2\hbar^2}{2mL^2} + \frac{1}{3}\frac{9\pi^2\hbar^2}{2mL^2}\\ &= \frac{14\pi^2\hbar^2}{6mL^2}\\ &= \frac{14}{3} E_1 \end{align*}
What is the quantum state of this particle at some later time \(t\)?
To find the time evolved state, stick a complex, time-dependent phase onto each term of the energy eigenstate expansion of the state (which we're given!)
\begin{align*} \left|{\Psi(t)}\right\rangle &= \frac{1}{\sqrt{3}}\big(e^{\frac{-iE_1t}{\hbar}}\left|{\phi_1}\right\rangle -e^{\frac{-iE_2t}{\hbar}}\left|{\phi_2}\right\rangle +ie^{\frac{-iE_3t}{\hbar}}\left|{\phi_3}\right\rangle \big)\\ &= \frac{1}{\sqrt{3}}\big(e^{\frac{-i\pi^2\hbar t}{2mL}}\left|{\phi_1}\right\rangle -e^{\frac{-i4\pi^2\hbar t}{2mL}}\left|{\phi_2}\right\rangle +ie^{\frac{-i9\pi^2\hbar t}{2mL}}\left|{\phi_3}\right\rangle \big)\\ \end{align*}
At some later time t, write an expression (do not evaluate) for the probability that a position measurement yields a result in the first half of the well, \(0<x<\frac{L}{2}\).
The probability of obtaining a position in the first half of the well is \[ P\left(0<x<\frac{L}{2}\right)=\int_{0}^{L/2} |\psi(x,t)|^2\,dx, \] where \(\psi(x,t)=\langle x|\Psi(t)\rangle\). Since \[ |\Psi(t)\rangle=\frac{1}{\sqrt{3}}\Big(e^{-iE_1t/\hbar}|\phi_1\rangle-e^{-iE_2t/\hbar}|\phi_2\rangle+i\,e^{-iE_3t/\hbar}|\phi_3\rangle\Big), \] we have \[ \psi(x,t)=\frac{1}{\sqrt{3}}\Big(e^{-iE_1t/\hbar}\phi_1(x)-e^{-iE_2t/\hbar}\phi_2(x)+i\,e^{-iE_3t/\hbar}\phi_3(x)\Big), \] where \(\phi_n(x)=\sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right)\) and \(E_n=\dfrac{n^2\pi^2\hbar^2}{2mL^2}\)
(No evaluation is required.)
At time \(t=\hbar/E_1\), if the energy of the particle is measured, what possible energy values may be obtained, and with what probability would each possible value be obtained? Check Beasts: Verify that \(\hbar/E_1\) is a time.
Since the Hamiltonian is not time-dependent, then the probabilities of measuring certain energy value will not depend on time!
So, my answers are the same as for part (b).