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#3 Solutions

  • Page ID
    92310
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    Q1

    Starting with the ansatz wavefunction in Equation 3.1.18, derive the time-independent Schrödinger equation from the more general time-dependent Schrödinger equation.

    Start with the ansatz wavefunction:

    \[\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt / \hbar}\tag{1}\]

    And insert it into the time-dependent Schrödinger:

    \[{ i\hbar\dfrac{\partial}{\partial t}\Psi(\vec{r},t)=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\Psi(\vec{r},t)}\tag{2}\]

    \[{ i\hbar\dfrac{\partial}{\partial t}\psi(\vec{r})e^{-iEt / \hbar}=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})e^{-iEt / \hbar}}\tag{3}\]

    Now take the time derivative on the left hand side:

    \[{ i\hbar\psi(\vec{r})\dfrac{\partial}{\partial t}e^{-iEt / \hbar}=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})e^{-iEt / \hbar}}\tag{4}\]

    \[{ i\hbar\psi(\vec{r})\dfrac{-iE}{\hbar}e^{-iEt / \hbar}=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})e^{-iEt / \hbar}}\tag{5}\]

    Elminate the exponential from both sides and simplify

    \[{ \psi(\vec{r})\dfrac{-i^2E\hbar}{\hbar}=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})}\tag{6}\]

    To recover the time-independent Schrödinger equation:

    \[{ E\psi(\vec{r})=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})}\tag{7}\]

    Q2

    Write the Schrödinger equation for a particle of mass m moving in a 2-dimensional space with the potential energy given by \(V(x, y) = -\dfrac {(x^2 + y^2)}{2}\)

    We need to take the time-independent version and express it in 2-d with the proper potential:

    \[{ \left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{1}\]

    Write \(\nabla^2\) in 2-d:

    \[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2})+V(\vec{r})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{2}\]

    Insert the potential:

    \[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2})+(-\dfrac {(x^2 + y^2)}{2})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{3}\]

    And finally change the functions to 2-d.

    \[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2})-\dfrac {(x^2 + y^2)}{2}\right]\psi(x,y) = E\psi(x,y)}\tag{4}\]

    Q3

    Demonstrate that the function \(e^{ikx}\) is an eigenfunction of either momentum operator.

    To be an eigenfunction when operated on the function must return itself and a constant:

    \[\hat{A}\psi = k \psi\tag{1}\]

    The linear momentum operator is:

    \[\hat p_x = -i \hbar \dfrac{\partial}{\partial x} \tag{2}\]

    Check the linear momentum operator

    \[\hat p_xe^{ikx} = -i \hbar \dfrac{\partial}{\partial x}e^{ikx} \tag{3}\]

    Take the derivative

    \[\hat p_xe^{ikx} = -i \hbar ike^{ikx} \tag{4}\]

    Simplify

    \[\hat p_xe^{ikx} = \hbar k e^{ikx} \tag{5}\]

    Which gives an eigenvalue of \(\hbar k\)

    Q4

    Consider a particle of mass \(m\) in a one-dimensional box of width \(L\). Calculate the general formula for energy of the transitions between neighboring states. How much energy is required to excite the particle from the \(n=1\) to \(n=2\) state?

    \(E_{n+1}- E_n= \dfrac{(n+1)^{2}h^{2}}{8mL^{2}}-\dfrac{n^{2}h^{2}}{8mL^{2}}=\dfrac{(n^2+2n+1)h^{2}}{8mL^{2}}-\dfrac{n^{2}h^{2}}{8mL^{2}}=\dfrac{(2n+1)h^{2}}{8mL^{2}}, n=1,2,....\)

    or

    \(E_n- E_{n-1}= \dfrac{n^{2}h^{2}}{8mL^{2}}-\dfrac{(n-1)^{2}h^{2}}{8mL^{2}}=\dfrac{n^{2}h^{2}}{8mL^{2}}-\dfrac{(n^2-2n+1)h^{2}}{8mL^{2}}=\dfrac{(2n-1)h^{2}}{8mL^{2}}, n=2,3,....\)

    The energy required to excite the particle from the \(n=1\) to \(n=2\) state is \(\dfrac{3h^{2}}{8mL^{2}}\).

    *The coefficients may vary since \(\dfrac{h^{2}}{8m}=\dfrac{(2\pi\hbar)^{2}}{8m}=\dfrac{4\pi^{2}\hbar^{2}}{8m}=\dfrac{\pi^{2}\hbar^{2}}{2m}\).

    Q5

    Consider a particle of mass \(m\) in a two-dimensional square box with sides \(L\). Calculate the four lowest (different!) energies of the system. Write them down in the increasing order with their principal quantum numbers. The energies of particle of mass \(m\) in a two-dimensional square box with sides \(L\) are

    \(E_{n_x,n_y}=\dfrac{h^{2}}{8mL^{2}}({n_x^{2}}+{n_y^{2}})\), \(n_x,n_y=1,2,...\).

    The four lowest (different!) energies of the system in the increasing order with their principal quantum numbers:

    \(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{1^{2}})=\dfrac{2h^{2}}{8mL^{2}}=\dfrac{h^{2}}{4mL^{2}}\) \(n_x,n_y=1,1\)

    \(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{2^{2}})=\dfrac{5h^{2}}{8mL^{2}}\) \(n_x,n_y=1,2\) or \(2,1\) (degenerate state)

    \(\dfrac{h^{2}}{8mL^{2}}({2^{2}}+{2^{2}})=\dfrac{8h^{2}}{8mL^{2}}=\dfrac{h^{2}}{mL^{2}}\) \(n_x,n_y=2,2\)

    \(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{3^{2}})=\dfrac{10h^{2}}{8mL^{2}}=\dfrac{5h^{2}}{4mL^{2}}\) \(n_x,n_y=1,3\) or \(3,1\) (degenerate state)

    *The coefficients may vary since \(\dfrac{h^{2}}{8m}=\dfrac{(2\pi\hbar)^{2}}{8m}=\dfrac{4\pi^{2}\hbar^{2}}{8m}=\dfrac{\pi^{2}\hbar^{2}}{2m}\).

    Q6

    The energies of a particle of mass \(m\) in a two-dimensional rectangular box with dimensions \(L_x\) and \(L_y\), where \(\dfrac{L_x}{L_y}=2\), are

    \[E_{n_x,n_y}=\dfrac{h^{2}}{8m}\left(\dfrac{n_x^{2}}{L_x^{2}}+\dfrac{n_y^{2}}{L_y^{2}}\right)\]

    with \(n_x,n_y=1,2,.... \infty\)

    The four lowest energies of the system in the increasing order with their principal quantum numbers:

    \[\dfrac{h^{2}}{8m}\left(\dfrac{1}{L_x^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{1}{4L_y^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{1}{4L_y^{2}}+\dfrac{4}{4L_y^{2}}\right)=\dfrac{h^{2}}{8m}\dfrac{5}{4L_y^{2}}=\dfrac{5h^{2}}{32mL_y^{2}}\]

    with \(n_x,n_y=1,1\)

    \[\dfrac{h^{2}}{8m}\left(\dfrac{4}{L_x^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{4}{4L_y^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{4}{4L_y^{2}}+\dfrac{4}{4L_y^{2}}\right)=\dfrac{h^{2}}{8m}\dfrac{8}{4L_y^{2}}=\dfrac{8h^{2}}{32mL_y^{2}}\]

    with \(n_x,n_y=2,1\)

    \[\dfrac{h^{2}}{8m}\left(\dfrac{9}{L_x^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{9}{4L_y^{2}}+\dfrac{1}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{9}{4L_y^{2}}+\dfrac{4}{4L_y^{2}}\right)=\dfrac{h^{2}}{8m}\dfrac{13}{4L_y^{2}}=\dfrac{13h^{2}}{32mL_y^{2}}\]

    with \(n_x,n_y=3,1\)

    \[\dfrac{h^{2}}{8m}\left(\dfrac{1}{L_x^{2}}+\dfrac{4}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{1}{4L_y^{2}}+\dfrac{4}{L_y^{2}}\right)=\dfrac{h^{2}}{8m}\left(\dfrac{1}{4L_y^{2}}+\dfrac{16}{4L_y^{2}}\right)=\dfrac{h^{2}}{8m}\dfrac{17}{4L_y^{2}}=\dfrac{17h^{2}}{32mL_y^{2}}\]

    with \(n_x,n_y=1,2\)

    *The coefficients may vary since \[\dfrac{h^{2}}{8m}=\dfrac{\left(2\pi\hbar\right)^{2}}{8m}=\dfrac{4\pi^{2}\hbar^{2}}{8m}=\dfrac{\pi^{2}\hbar^{2}}{2m}\]

    Q7

    The expectation value of the position operator is:

    \[\langle \hat{x}\rangle = \int_{0}^{a}\psi_n^*\hat{x}\psi_n dx\tag{1}\]

    \[\langle \hat{x}\rangle = \dfrac{2}{a}\int_{0}^{a}x\sin^2\dfrac{\pi x}{a}dx = \dfrac{a}{2}\tag{2}\]

    Where we used the half angle formula and integration by parts to compute the integral.

    Q8

    The expectation value of the position operator squared is:

    \[\langle \hat{x^2}\rangle = \int_{0}^{a}\psi_n^*\hat{x^2}\psi_n dx\tag{3}\]

    \[\langle \hat{x^2}\rangle = \dfrac{2}{a}\int_{0}^{a}x^2\sin^2\dfrac{\pi x}{a}dx = \dfrac{a^2}{3} - \dfrac{a^2}{2\pi^2}\tag{4}\]

    Again we used the half angle formula and integration by parts to compute this integral.

    Q9

    The expectation value of the momentum operator is:

    \[\langle \hat{p}\rangle = \int_{0}^{a}\psi_n^*\hat{p}\psi_n dx\tag{5}\]

    \[\langle \hat{p}\rangle = -i\hbar\dfrac{\pi}{a}\dfrac{d}{dx}\tag{6}\]

    \[\langle \hat{p}\rangle = \dfrac{2}{a}\dfrac{-i\hbar \pi}{a}\int_{0}^{a}\sin\dfrac{\pi x}{a}\cos\dfrac{\pi x}{a}dx = 0\tag{7}\]

    The integral in Equation 7 is zero because sin(x) times cos(x) is an odd function and the integral of an odd function over a symmetric interval is zero.

    Q10

    The expectation value of the momentum operator squared is:

    \[\langle \hat{p^2}\rangle = \int_{0}^{a}\psi_n^*\hat{p^2}\psi_n dx\tag{8}\]

    \[ \hat{p^2} = -\hbar^2\dfrac{\pi^2}{a^2}\dfrac{d^2}{dx^2}\tag{9}\]

    \[\langle \hat{p^2}\rangle = \dfrac{2}{a}\dfrac{-\hbar^2 \pi^2}{a^2}\int_{0}^{a}\sin^2\dfrac{\pi x}{a}dx = \hbar^2\dfrac{\pi^2}{a^2}\tag{10}\]

    \[\Delta p = \sqrt{\langle \hat{p^2}\rangle - \langle \hat{p}\rangle^2 } \tag{11}\]

    \[\Delta x = \sqrt{\langle \hat{x^2}\rangle - \langle \hat{x}\rangle^2 } \tag{12}\]

    \[\Delta p\Delta x = \left(\hbar\dfrac{\pi}{a}\right) \left(\dfrac{a^2}{3} - \dfrac{a^2}{2\pi^2} - \dfrac{a^2}{4}\right)^\dfrac{1}{2}\tag{13}\]

    \[(\dfrac{\hbar}{2})(\dfrac{\pi^2}{3} - 2)^\dfrac{1}{2} = 1.14*\dfrac{\hbar}{2}\tag{14}\]

    Thus \(\Delta x\) times \(\Delta p\) is 1.14 times greater than \(\dfrac{\hbar}{2}\) and the Heisenberg Uncertainty principle holds.

    Q11

    Solve by calculation of the expectation value using the proper particle in the box wave functions.

    \[\langle x \rangle = \int_{-\infty}^\infty \psi_n^*(x) \hat{x}\ \psi_n(x) dx\tag{1}\]

    Input proper particle in the box \(\psi\) for n=1

    \[\langle x \rangle = \int_{0}^L \sqrt{\dfrac{2}{L}}sin{\dfrac{\pi x}{L}} \hat{x}\ \sqrt{\dfrac{2}{L}}sin{\dfrac{\pi x}{L}} dx\tag{2}\]

    Simplify knowing \(\hat{x}\) means multiply by \(x\).

    \[\langle x \rangle = \dfrac{2}{L} \int_{0}^L x sin^2 {\dfrac{\pi x}{L}} dx\tag{3}\]

    Integrate

    (This is not a math class. If you have problems doing this integral try using the Half-Angle Identities. On a test you will have a table of integrals.)

    \[\langle x \rangle = \dfrac{L}{2}\tag{4}\]

    Now for n=2

    \[\langle x \rangle = \dfrac{2}{L} \int_{0}^L x sin^2 {\dfrac{2 \pi x}{L}} dx\tag{5}\]

    Integrate
    \[\langle x \rangle = \dfrac{L}{2}\tag{6}\]

    These answers make sense because the particle must be in the box and it has an equal chance of being on the right side or the left side, thus the average is the middle.

    170px-Particle_in_a_box_wavefunctions_2.svg.png

    Q12

    The most probable position is where the amplitude of the probability is the largest.

    Mathematically you can do this by

    \[\dfrac{d}{dx} \psi_n(x) = 0 \tag{7}\]

    \[\dfrac{d}{dx} sin{\dfrac{\pi x}{L}} = 0\tag{8}\]

    \[\dfrac{d}{dx} sin{\dfrac{2 \pi x}{L}} = 0 \tag{9}\]

    and solving for x using n=1 and n=2 respectively.

    The easy way is just to note that the maximum for n=1 occurs in the middle of the box so the most probable position is

    \[x = \dfrac{L}{2}\tag{10}\]

    For n=2 there are two "maximums" one positive and one negative. These occur at

    \[x = \dfrac{L}{4}, \dfrac{3L}{4} \tag{11}\]

    Q13

    \[\Delta x = \sqrt{\langle \hat{x^2}\rangle - \langle \hat{x}\rangle^2 } \tag{12}\]

    See Q7 and Q8 for finding \(\langle \hat{x^2}\rangle\) and \(\langle \hat{x}\rangle^2\).

    \[\Delta x =\sqrt{\dfrac{a^2}{3} - \dfrac{a^2}{2\pi^2} - \dfrac{a^2}{4}}=a\sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} }=0.18a\]

    Q14

    They should be normalized so that the interpretation of probability as the wave function squared makes sense. The probability of finding the particle somewhere must be 1 or 100% so the wave functions are scaled correctly with a normalization constant.

    With unnormalized wave functions an expectation value will be incorrect.

    \[ \langle x \rangle = \int_{-\infty}^{+\infty} \psi^*(x) x \psi(x) dx = \dfrac{L^2}{4} \tag{12} \]

    Add the normalization constant:

    \[ \langle x \rangle = \dfrac{\int_{-\infty}^{+\infty} \psi^*(x) x \psi(x) dx}{\int_{-\infty}^{+\infty} \psi^*(x)\psi(x) dx} = \dfrac{L^2}{4}\dfrac{2}{L} = \dfrac{L}{2} \tag{13} \]


    #3 Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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