Skip to main content
Chemistry LibreTexts

#4 Solution

  • Page ID
    92312
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Q1

    Use the normalization condition to find \(A\) in the following wavefunction:

    \[ \psi(x) = Ae^{ikx}e^{-x^2/2a^2}\]

    Solution 1:

    Normalization condition is \(\int_{-\infty}^{\infty} \psi^*(x) \psi(x)dx=1\). To calculate this integral we need \(\psi^*(x)\) which we find by taking complex conjugate of \(\psi(x)\). Taking complex conjugate of \(\psi(x)\) in this case is simply reversing the sign before \(i\) : \( \psi^*(x) = Ae^{-ikx}e^{-x^2/2a^2}\). Further we plug \(\psi(x)\) and \(\psi^*(x)\) in the normalization condition:

    \[ \int_{-\infty}^{\infty} \psi^*(x) \psi(x)dx=\int_{-\infty}^{\infty} Ae^{-ikx}e^{-x^2/2a^2}Ae^{ikx}e^{-x^2/2a^2}dx=\int_{-\infty}^{\infty} A^{2}e^{-x^2/2a^2}e^{-x^2/2a^2}dx=\\ \int_{-\infty}^{\infty} A^{2}e^{-2x^2/2a^2}dx=\int_{-\infty}^{\infty} A^{2}e^{-x^2/a^2}dx= a/a \int_{-\infty}^{\infty} A^{2}e^{-(x/a)^2} dx=\\ a \int_{-\infty}^{\infty} A^{2}e^{-(x/a)^2}(\dfrac{1}{a})dx = a \int_{-\infty}^{\infty} A^{2}e^{-(x/a)^2}d(\dfrac{x}{a})=\text (substitution \ of \ variables)=A^{2}a\sqrt{\pi}=1\]

    \[A=\pm\sqrt{1/(a\sqrt{\pi})}\]

    Q2

    Find the normalization constant for the wavefunction:

    \[
    \psi(x)=\left\{
    \begin{array}{ll}
    C \quad \frac{-c}{4} \leq x \leq \frac{c}{4} \\
    0 \qquad \textrm{elsewhere} \\
    \end{array}
    \right.
    \]

    Solution 2:

    Assume the normalization constant is A. Use the normalization condition:

    \[\int_{-\infty}^{\infty} A\Psi ^{*} (x) A\Psi (x) dx = 1\]

    \[1= \int_{-\infty}^{\infty} A\Psi ^{*} (x) A\Psi (x) dx = A^2\int_{-\infty}^{\infty}\Psi^* \Psi(x) dx=A^2 \int_{-\infty}^{-\dfrac{C}{4}} 0\times0 dx +A^2 \int_{\dfrac{C}{4}}^{\infty} 0\times0 dx +A^2 \int_{-\dfrac{C}{4}}^{\dfrac{C}{4}} C^2 dx=0+0+A^2 C^2 \dfrac{C}{2}=1\]

    So \(A=\pm \sqrt{\dfrac{2}{C^3}}\)

    Q3

    Show that the eigenstates to a particle in a 1D box with infinite potential satify the orthogonormality relationship:

    \[ \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m(x)\; dx = 1 \; \text{if } m=n\]

    and

    \[ \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m(x)\; dx = 0 \; \text{if } m \neq n\]

    What is the origin of the first equation?

    Solution 3:

    Eigenstates of a particle in 1D box of length \(L\) is given by the expression \(\psi_n = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi x}{L}}, n=1,2,...\). To show that these eigenstates satisfy \((1)\) we need to plug \(\psi_n\) with the same principal quantum number \(n\) (or subscript) in expression \((1)\):

    \[ \int_{0}^{L} \psi_n^*(x) \psi_n(x) dx =\int_{0}^{L} \dfrac{2}{L}\sin^2{\dfrac{n\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{0}^{L}(1-cos{\dfrac{2n\pi x}{L}})dx=\\ \dfrac{1}{L}x\Biggr\rvert_{0}^{L} +\dfrac{1}{L}\dfrac{L}{2n\pi}(-\sin{\dfrac{2n\pi x}{L}})\Biggr\rvert_{0}^{L}=\dfrac{1}{L}L-\dfrac{1}{L}0-\sin{\dfrac{2n\pi L}{L}}+\sin{(0)}=1-0-0+0=1\]

    This is correct according to Postulate 2 (normalization condition).

    To show that the eigenstates satisfy \((2)\) we need to plug \(\psi_n\) and \(\psi_m\) with different principal quantum numbers \(n\) and \(m\) (or subscripts) in expression \((2)\):

    \[ \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m(x) dx =\int_{0}^{L} \dfrac{2}{L}\sin{\dfrac{m\pi x}{L}}\sin{\dfrac{n\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{0}^{L}\cos{(\dfrac{m\pi x}{L}-\dfrac{n\pi x}{L})}-\cos{(\dfrac{m\pi x}{L}+\dfrac{n\pi x}{L})}dx= \\ \dfrac{1}{L}\int_{0}^{L}\cos{\dfrac{(m-n)\pi x}{L}}+\cos{\dfrac{(m+n)\pi x}{L}}dx=\dfrac{1}{L}(\dfrac{L}{(m-n)\pi}\sin{\dfrac{(m-n)\pi x}{L}}+\dfrac{L}{(m+n)\pi}\sin{\dfrac{(m+n)\pi x}{L}})\Biggr\rvert_{0}^{L}=\\ \dfrac{1}{L}(\dfrac{L}{(m-n)\pi}\sin{\dfrac{(m-n)\pi L}{L}}-0+\dfrac{L}{(m+n)\pi}\sin{\dfrac{(m+n)\pi L}{L}}-0)=\\ \dfrac{1}{L}(\dfrac{L}{(m-n)\pi}\sin{((m-n)\pi)}-0+\dfrac{L}{(m+n)\pi}\sin{((m+n)\pi)}-0) =\dfrac{1}{L}(0-0+0-0)=0\]

    The origin of the first equation is the same as the origin of the normalization condition. By stating \((1)\) or normalization condition we mean that the object that we are describing with the wavefunctions \({\psi_n(x)}\) exists.

    Q4

    A particle of mass \(m\) moves in a one-dimensional box of length \(L\) with boundaries at \(x = 0\) and \(x = L\).

    • For any generic wavefunction \(\psi_n(x)\) for this system, calculate the probability that the particle is found somewhere in the region 0 ≤ x ≤ L/4.
    • Show how this probability depends on \(n\).
    • For what value of n is there the largest probability of finding the particle in 0 ≤ x ≤ L/4?

    Solution 4

    \[ P(0 ≤ x ≤ L/4)=\int_{0}^{L/4} \psi^*(x) \psi(x)dx=\int_{0}^{L/4}\dfrac{2}{L}\sin^2{\dfrac{n\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{0}^{L/4}(1-\cos{\dfrac{2n\pi x}{L}})dx=\\ \dfrac{1}{L}x\Biggr\rvert_{0}^{L/4} +\dfrac{1}{L}\dfrac{L}{2n\pi}(-\sin{\dfrac{2n\pi x}{L}})\Biggr\rvert_{0}^{L/4}=\dfrac{1}{L}\dfrac{L}{4}-\dfrac{1}{L}0-\dfrac{1}{2n\pi}\sin{\dfrac{2n\pi L}{4L}}+\dfrac{1}{2n\pi}\sin{(0)}=\\ \dfrac{1}{4}-0-\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}+0=\dfrac{1}{4}-\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}\]

    \(\dfrac{1}{4}-\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}\) is the largest when \(\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}\) is the smallest. Since the latter is the product of the oscillating \(\sin{\dfrac{n\pi}{2}}\) and decreasing \(\dfrac{1}{2n\pi}\) function we can say that there is no smallest value of \(\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}\). However, when \(n\) is getting larger \(\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}}\) is getting smaller (see figure below). So, the answer is \(n=\infty\).

    Q5

    Evaluate the following commutators \( [\hat{A}, \hat{B}] \) for the following pair of operators:

    1. \(\hat{C}\) and \([\hat{D},\hat{C}]\hat{E} \)
    2. \(\frac{d}{dy}-y\) and \(\frac{d}{dy}+y \)
    3. \(\frac{d}{dy}\) and \(\int_{0}^{y} dy \)
    4. \( \frac{d^2}{dy^2}\) and \(y \)
    5. \(2\) and \(\frac{d}{dy} \)

    Solution 5

    a.

    \(\hat{A} =\hat{C}\), \(\hat{B}=[\hat{D},\hat{C}]\hat{E}

    \[[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}=\hat{C}[\hat{D},\hat{C}]\hat{E}-[\hat{D},\hat{C}]\hat{E}\hat{A}=\hat{C}(\hat{D}\hat{C}-\hat{C}\hat{D})\hat{E}-(\hat{D}\hat{C}-\hat{C}\hat{D})\hat{E}\hat{A}=\hat{C}\hat{D}\hat{C}\hat{E}-\hat{C}^2 \hat{D}\hat{E}-\hat{D}\hat{C}\hat{E}\hat{A}+\hat{C}\hat{D}\hat{E}\hat{A} \]

    b.

    \(\hat{A}=\dfrac{d}{dy}-y\), \(\hat{B}=\dfrac{d}{dy}+y\)

    Assume that we operate \([\hat{A},\hat{B}]\) on a random function f(y)

    So \[ [\hat{A},\hat{B}] f(y) =\hat{A}\hat{B}f(y)-\hat{B}\hat{A}f(y) = (\dfrac{d}{dy}-y)(\dfrac{d}{dy}+y)f(y) -(\dfrac{d}{dy}+y)(\dfrac{d}{dy}-y)f(y)\]

    \[=(\dfrac{d}{dy}-y)\left[\dfrac{df}{dy}+yf(y)\right]-(\dfrac{d}{dy}+y)\left[\dfrac{df}{dy}-yf(y)\right]=\dfrac{d^2f}{dy^2}+\dfrac{d}{dy}[y(f(y)]-y\dfrac{df}{dy}-y^2f(y)-\left[\dfrac{d^2f}{dy^2}-\dfrac{d}{dy}(yf(y))+y\dfrac{df}{dy}-y^2f(y)\right]\]

    \[=\dfrac{d^2f}{dy^2}+f(y)+y\dfrac{df}{dy}-y\dfrac{df}{dy}-y^2f(y)-\dfrac{d^2f}{dy^2}+f(y)+y\dfrac{df}{dy}-y\dfrac{df}{dy}+y^2f(y)=2f(y) \]

    So \( [\hat{A},\hat{B}]=2\)

    (Note: pay attention to the order of differentiation! It is always safe to operate on a random function to make sure your results are correct.)

    c.

    \(\hat{A}=\dfrac{d}{dy}\), \(\hat{B}=\int_{0}^{y}dy\)

    For \(\hat{B}\),

    \(\hat{B}f(y)=\int_{0}^{y}f(y)dy\).

    So,

    \[[\hat{A},\hat{B}]f(y)=\dfrac{d}{dy}\int_{0}^{y}f(y)dy-\int_{0}^{y}\dfrac{df(y)}{dy}dy=f(y)-\left[f(y)-f(0)\right]=f(0)\]

    d.

    \[[\hat{A},\hat{B}]f(y)=\dfrac{d^2}{dy^2}\left[yf(y)\right]-y\dfrac{d^2f}{dy^2}=\dfrac{d}{dy}\left[f(y)+y\dfrac{df}{dy}\right]-y\dfrac{d^2f}{dy^2}\]

    \[=\dfrac{df}{dy}+\dfrac{df}{dy}+y\dfrac{d^2f}{dy^2}-y\dfrac{d^2f}{dy^2}=2\dfrac{df}{dy}=2\dfrac{d}{dy}f(y)\]

    So \([\hat{A},\hat{B}]=2\dfrac{d}{dy}\)

    e.

    \[[\hat{A},\hat{B}]f(y)=2\dfrac{d}{dy}f(y)-\dfrac{d}{dy}\left[2f(y)\right]=2\dfrac{df}{dy}-2\dfrac{df}{dy}=0\]

    So \[[\hat{A},\hat{B}]=0\]

    (This means \(\hat{A}\) and \(\hat{B}\) commute.)

    Q6

    For the following molecules identify the number of

    • degrees of freedom,
    • translational degrees of freedom
    • rotational degrees of freedom
    • vibrational degrees of freedom
    1. \(SF_6\)
    2. \(CO_2\)
    3. \(O_2\)
    4. \(C_{60}\)
    5. \(Ar\)

    Solution 6

    degrees of freedom = translational degrees of freedom + rotational degrees of freedom + vibrational degrees of freedom

    The total number of degrees of freedom is equal to 3n, where n is the number of particles in a body. The number of translational degrees of freedom is equal to 3. The number of rotational degrees of freedom is equal to 3 for nonlinear objects and 2 for linear objects. The number of vibrational degrees of freedom is equal to 3n-6 (which is 3n-3-3) for nonlinear objects and 3n-5 (which is 3n-3-2) for linear objects.

    Molecule # of degrees of freedom (df) # of translational df # of rotational df # of vibrational df
    \(SF_6\) 3x7=21 3 3 21-3-3=15
    \(CO_2\) 3x3=9 3 2 9-3-2=4
    \(O_2\) 3x2=6 3 2 6-3-2=1
    \(C_{60}\) 3x60=180 3 3 180-3-3=174
    \(Ar\) 3x1=3 3 0 0

    Q7

    At what point(s) during the oscillation of a spring that obey's Hooke's law is the force on the mass the greatest?

    Solution 7

    Recall that F = - kx . Thus the force on the mass will be greatest when the displacement of the block is at its maximum, or when x = ±x m .

    Q8

    The wavefunctions for the quantum mechanical harmonic oscillator \(| \psi_v \rangle\) in atomic units with \(\alpha = 1 \) are expressed as

    \[ | \psi_v \rangle = N_v H_v e^{-(x-x_o)^2/2} \]

    with

    • \(x_o\) is the equilibrium position of the oscillator
    • \(N_v\) is a normalization factor for a specific \(v\) value
    • \(H_v\) is the Hermite polynomial for a specific \(v\) value (see Table M1)

    What is the wavefunction (with determined normalization factor) for the harmonic oscillator in the \(v=0\) state?

    Solution 8

    To write out the full wave function we need to find H0 and N0, the table of H values can be found in the lecture notes with H0 equals one. To calculate N0 we have to normalize \(| \psi_0 \rangle\) .

    \[ \int_{-\infty}^{\infty} \psi_0^*(x) \psi_0(x)\ dx = 1 \tag{1}\]

    \[ N_0^2 \int_{-\infty}^{\infty} e^{-(x-x_0)^2}dx = 1 \tag{2}\]

    We use the substitution y = x-x0

    \[ N_0^2 \int_{-\infty}^{\infty} e^{-(y)^2}dy = 1 \tag{3}\]

    \[N_0^2\pi^{1/2} = 1 \tag{4}\]

    \[N_0 = \pi^{-1/4} \tag{5}\]

    \[ | \psi_0 \rangle = \pi^{-1/4} e^{-(x-x_0)^2/2} \tag{6}\]

    Q9

    Calculate the mean displacement of the oscillator \(x\) from equilibrium when the oscillator is in the \(v=0\) and \(v=1\) quantum states? Explain the differences in your own words.

    Solution 9

    For the \(v=0\) quantum state the wavefunction and mean displacement are:

    \[ | \psi_0 \rangle = \pi^{-1/4} e^{-(x-x_o)^2/2} \tag{7}\]

    \[<\hat{x}>_0 = (\frac{1}{\pi})^\frac{1}{2}\int_{-\infty}^{\infty} (x-x_o)e^{-(x-x_o)^2}dx = 0\tag{8}\]

    Substitue \(y=x-x_o\),

    \[<\hat{x}>_0 = (\frac{1}{\pi})^\frac{1}{2}\int_{-\infty}^{\infty} ye^{-(y)^2}dy = 0\tag{8}\]

    This integral is equal to zero because we are integrating an odd function over symmetric limits (\(y\) is an odd function while \(e^{y^2}\) is an even function and an even function times an odd function is odd).

    For the \(v=1\) quantum the wavefunction and mean displacement are:

    \[ | \psi_1 \rangle = (\frac{4}{\pi})^\frac{1}{4} 2xe^{-(x)^2/2} \tag{9}\]

    \[<\hat{x}>_1 = 4*(\frac{4}{\pi})^\frac{1}{2}\int_{-\infty}^{\infty} x^3e^{-(x)^2}dx = 0\tag{10}\]

    Again the integral is zero because we are integrating an odd function over symmetric limits.

    As they are both zero there is no difference between the mean displacements because both are symmetric around zero.

    Q10

    Calculate the mean displacement squared (\(\langle x^2 \rangle\)) of the oscillator \(x\) from equilibrium when the oscillator is in the \(v=0\) and \(v=1\) quantum states? Explain the differences in your own words.

    Solution 10

    Using the same \(| \psi_0 \rangle\) and \(| \psi_1 \rangle\) from the previous problem the mean displacements squared are:

    \[<\hat{x}^2>_1 = 4*(\frac{4}{\pi})^\frac{1}{2}\int_{-\infty}^{\infty} x^4e^{-(x)^2}dx = 4*(\frac{4}{\pi})^\frac{1}{2} \times \frac{3\pi^\frac{1}{2}}{4} =6\tag{11}\]

    \[<\hat{x}^2>_0 = (\frac{1}{\pi})^\frac{1}{2}\int_{-\infty}^{\infty} x^2e^{-(x)^2}dx = (\frac{1}{\pi})^\frac{1}{2} \times \frac{\pi^\frac{1}{2}}{2} = \frac{1}{2}\tag{12}\]

    The difference in the mean displacement squared is due to the probability density of \(| \psi_1 \rangle\) being wider than the probability density of \(| \psi_0 \rangle\) as can be seen on the graphs in the lecture notes.

    Q11

    Use the answers from Q9 and Q10 to determined the uncertainty of position of a particle following the harmonic oscillator with \(v=0\)?

    Solution 11

    \[\Delta x_0 = (<x^2> - <x>^2)^{1/2}\]

    \[\Delta x_0 = {1/2}^{1/2}\]

    Q12

    For a harmonic oscillator with a mass of \(1.21 \times 10^{-25} \;kg\), the energy levels are separated by \(4.82 \times 10^{-21}\; J\). What is the force constant for the oscillator? What is the zero point energy of this oscillator?

    Solution 12

    \[\Delta E = 4.82 \times 10^{-21}\; J\]

    \[\Delta E = \hbar\omega\]

    \[\omega=\dfrac{\Delta E}{\hbar}\]

    \[\omega = (K/m)^{1/2}\]

    \[K = m\omega^2=m\left(\dfrac{\Delta E}{\hbar}\right)^2\]

    \[K = 252.47\; N/M\]

    The zero point energy, or the lowest possible energy for the harmonic oscillator is:

    \[E = 1/2\hbar\omega\]

    \[E = 1/2\Delta E\]

    \[E = 2.41 \times 10^{-21}\; J\]


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

    • Was this article helpful?