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#7A Solutions

  • Page ID
    92320
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    Q7.1

    What is the orbital angular momentum of an electron in the following orbitals

    1. 1s
    2. 2s
    3. 2p
    4. 3d
    5. 5f

    How many angular and radial nodes exist for the wavefunctions described by the above states?

    Solution 7.1

    Angular Momentum is

    \[L = \hbar \sqrt{ \ell(\ell +1)} \tag{24}\]

    And for the number of nodes

    Total Nodes \( = n -1 \)

    Angular Nodes \( = \ell \)

    Radial Nodes \( = n -1 - \ell \)

    1. 1s, \(L = 0\), Angular Nodes \( = 0 \), Radial Nodes \( = 0 \)
    2. 2s, \(L =0 \), Angular Nodes \( = 0 \), Radial Nodes \( = 1 \)
    3. 2p, \(L =\hbar \sqrt{2} \), Angular Nodes \( = 1 \), Radial Nodes \( = 0 \)
    4. 3d, \(L =\hbar \sqrt{6} \), Angular Nodes \( = 2 \), Radial Nodes \( = 0 \)
    5. 5f, \(L =\hbar \sqrt{12} \), Angular Nodes \( = 3 \), Radial Nodes \( = 1 \)

    Q7.2

    Use Slater's rules to calculate \(Z_{eff}\) and \(Z\) for

    1. The valence electron of the Neon atom
    2. The the innermost electron of Lithium atom
    3. The \(4s^2\) electrons of the Br atom
    4. The outermost electron of the row 2 elemental atom with the largest effective nuclear charge

    Solution 7.2

    1. The valence electron of the Neon atom

    (1s2) (2s2 2p6)

    S = 0.35x7+0.85x2 = 4.15

    Zeff = Z-S = 10-4.15 = 5.85

    b. The the innermost electron of Lithium atom

    (1s2) (2s1)

    S = 0.3x1 = 0.3

    Zeff = Z-S = 3-0.3 = 2.7

    c. The \(4s^2\) electrons of the Br atom

    (1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s2 4p5)

    S = 0.35x6+0.85x18+1.00x10 = 27.4

    Zeff = Z-S = 35-27.4= 7.6

    d. The outermost electron of the row 2 elemental atom with the largest effective nuclear charge

    Ne

    (1s2) (2s2 2p6)

    S = 0.35x7+0.85x2 = 4.15

    Zeff = Z-S = 10-4.15 = 5.85

    Q7.3

    Use the variational method with the unnormalized trial function

    \[ | \varphi \rangle = x(L−x) \rangle \]

    to estimate the ground state energy for a particle in a one-dimensional box of length \(L\). (Hint: If you want the general pain, you can find this answer in this paper, but it is not necessary)

    Solution 7.3

    From variational method:

    \[E_1 \leq \dfrac{\langle\varphi | H | \varphi\rangle}{\langle\varphi |\varphi\rangle} \]

    To get the upper limit for \(E_1\) from variational method we need to calculate the following integral:

    \[\dfrac{\langle\varphi| H | \varphi\rangle}{\langle\varphi |\varphi\rangle} \tag{1}\]

    First we must normalize \(\varphi\) by calculating the following integral:

    \[\langle\varphi | \varphi\rangle= \int_{0}^{L} \varphi(x)^*\varphi(x)dx = 1\]

    \[A^2\int_{0}^{L}x^2(L-x)^2dx = A^2\int_{0}^{L}(x^2 L^2 -2Lx^3 +x^4)dx = \\A^2(L^2\dfrac{x^3}{3}\Biggr\rvert_{0}^{L} - 2L\dfrac{x^4}{4} \Biggr\rvert_{0}^{L}+ \dfrac{x^5}{5}\Biggr\rvert_{0}^{L}) = A^2(\dfrac{L^5}{3} - \dfrac{2L^5}{4} + \dfrac{L^5}{5}) =A^2 \dfrac{(10-15+6)L^5}{30}\]

    \[A = \pm\sqrt{\dfrac{30}{L^5}} \]

    Then we calculate the numerator of \((1)\):

    \[\langle\varphi | H | \varphi\rangle = \int_{0}^{L}x(L−x) (- \dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}) x(L−x)dx = \\- \dfrac{\hbar^2}{2m} \int_{0}^{L} (xL - x^2) (-2) dx = \dfrac{\hbar^2}{m} (L\dfrac{x^2}{2} - \dfrac{x^3}{3}) \Biggr\rvert_{0}^{L} = \dfrac{\hbar^2}{m} (\dfrac{L^3}{2} - \dfrac{L^3}{3}) = \dfrac{\hbar^2}{m} \dfrac{L^3(3-2)}{6} = \dfrac{\hbar^2 L^3}{6m} \]

    Plug \((2)\) and \((3)\) in \((1)\):

    \[\dfrac{\langle\varphi| H | \varphi\rangle}{\langle\varphi |\varphi\rangle} = \dfrac{30}{L^5} \dfrac{\hbar^2 L^3}{6m} = \dfrac{5\hbar^2}{mL^2}\]

    The conclusion is:

    \[E_1 \leq \dfrac{5\hbar^2}{mL^2}\]

    Q7.4

    How would use the variational method approximation in Q2 to determined the energy of the next highest eigenstate for the particle in a box with \(n=2\)?

    Solution 7.4

    Variational method cannot be used for excited states. We need to use other theories (e.g., perturbation).

    Q7.5

    Consider the "quartic oscillator" with the following Hamiltonian

    \[ \hat{H} = \dfrac{1}{2} \dfrac{d^2}{dx^2} + \dfrac{1}{2} x^4\]

    1. What is the zero point energy of this this system determined with the variational method approximation using the unnormalized trial wavefunction \[ | \varphi \rangle = e^{-1/2 \alpha (x-x_o)^2} \]
    2. What is the value of \(\alpha\) for the trail wavefunction used in this approximation?
    3. How accurate would this wavefunction be in estimating the zero point energy of the harmonic oscillator?

    Solution 7.5

    a. What is the zero point energy of this this system determined with the variational method approximation using the unnormalized trial wavefunction \[ | \varphi \rangle = e^{-1/2 \alpha (x-x_o)^2} \]

    To simplify calculations we will use displacement coordinate \(q\) instead of \((x-x_o)\).

    Then, analogously to the solution in Q2:

    \[E_0 \leq \dfrac{\langle\varphi | H | \varphi\rangle}{\langle\varphi |\varphi\rangle} \]

    Normalization:

    \[\langle\varphi | \varphi\rangle= \int_{0}^{L} \varphi(q)^*\varphi(q)dq = 1\]

    \[A^2\int_{-\infty}^{\infty} e^{- \alpha q^2} dq = A^2 \sqrt{\dfrac{\pi}{\alpha}} = 1\]

    \[A = \pm(\dfrac{\alpha}{\pi})^{1/4} \]

    Numerator:

    \[\langle\varphi | H | \varphi\rangle = \int_{-\infty}^{\infty} e^{-\dfrac{1}{2} \alpha q^2} (\dfrac{1}{2} \dfrac{d^2}{dq^2} + \dfrac{1}{2} q^4) e^{-\dfrac{1}{2} \alpha q^2} dq = \\ \int_{-\infty}^{\infty} e^{-\dfrac{1}{2} \alpha q^2} (\dfrac{1}{2} (-\alpha + \alpha^2 q^2)e^{-\dfrac{1}{2} \alpha q^2} + \dfrac{1}{2} q^4 e^{-\dfrac{1}{2} \alpha q^2}) dq = \\ \int_{-\infty}^{\infty} e^{-\dfrac{1}{2} \alpha q^2} \dfrac{1}{2} (-\alpha + \alpha^2 q^2)e^{-\dfrac{1}{2} \alpha q^2} dq + \int_{-\infty}^{\infty} e^{-\dfrac{1}{2} \alpha q^2} \dfrac{1}{2} q^4 e^{-\dfrac{1}{2} \alpha q^2} dq = \\ \dfrac{1}{2} (\int_{-\infty}^{\infty} e^{-\alpha q^2} (-\alpha + \alpha^2 q^2) dq + \int_{-\infty}^{\infty} e^{-\alpha q^2} q^4 dq) = \\ \dfrac{1}{2} (-\alpha \int_{-\infty}^{\infty} e^{-\alpha q^2} dq + \alpha^2 \int_{-\infty}^{\infty} q^2 e^{-\alpha q^2} dq + \int_{-\infty}^{\infty} e^{-\alpha q^2} q^4 dq) = \\ \dfrac{1}{2} (-\alpha \sqrt{\dfrac{\pi}{\alpha}} + \alpha^2 \dfrac{1}{2}\sqrt{\dfrac{\pi}{ \alpha^3}} + \dfrac{3}{4} \sqrt{\dfrac{\pi}{\alpha^5}})\]

    \[\dfrac{\langle\varphi | H | \varphi\rangle}{\langle\varphi |\varphi\rangle} = \sqrt{\dfrac{\alpha}{\pi}} (- \dfrac{1}{2} \alpha \sqrt{\dfrac{\pi}{\alpha}} + \dfrac{1}{2} \alpha^2 \dfrac{1}{2} \sqrt{\dfrac{\pi}{\alpha^3}} + \dfrac{1}{2}\dfrac{3}{4} \sqrt{\dfrac{\pi}{\alpha^5}}= \\ -\dfrac{1}{2} \alpha + \dfrac{1}{4} \alpha + \dfrac{3}{8} \dfrac{1}{\alpha^2} = I \]

    b. What is the value of \(\alpha\) for the trail wavefunction used in this approximation?

    Optimal \(\alpha\) is found by taking the derivative of \(I\) and calculating the \(\alpha\) at which the derivative is equal to zero:

    \[ \dfrac{dI}{d \alpha} = -\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{3}{4} \dfrac{1}{\alpha^3} = 0 \]

    \[\alpha = - 3^{1/3}\]

    c. How accurate would this wavefunction be in estimating the zero point energy of the harmonic oscillator?

    The normalized trial wavefunction \( | \varphi \rangle = (\dfrac{\alpha}{\pi})^{1/4} e^{-1/2 \alpha (x-x_o)^2} \) has the same structure as the true wavefunction of the ground state of harmonic oscillator \(| \psi_0 \rangle = \pi^{-1/4} e^{-1/2(x-x_o)^2}\). The functions are equal when \(\alpha = 1\). Thus, the trial wavefunction would be accurate in estimating the zero point energy of the harmonic oscillator.

    Q7.6

    A basis function is an element of a particular basis for a function space. Every continuous function in the function space can be represented as a linear combination of basis functions, just as every vector in a vector space can be represented as a linear combination of basis vectors.

    • List three basis that can be used to expand a general 1-D function.
    • Write the mathematical expansion formula for each of the three expansions above

    Laplace Series Expansion

    \[f(x) = \sum_s^\infty A_s e^{-st}\]

    Fourier Series Expansion

    \[f(x) = a_0 + \sum_n^\infty a_ncosnx + b_nsinnx\]

    Taylor Series Expansion

    \[f(x) = \sum_n^\infty \frac{f^{(n)}(a)}{n!}(x - a)^n\]

    Q7.7

    What is the definition of a complete basis? Are the three basis systems identified above complete basis for describing 1-D functions?

    A set of unctions is considered complete if all other functions can be expressed in terms of them. All the expansions from part 1 are complete.

    Q7.8

    What are the following dot products for the particle in a box?

    1. \(\langle \psi_{n=1} | \psi_{n=1} \rangle \)
    2. \(\langle \psi_{n=1} | \psi_{n=2} \rangle \)
    3. \(\langle \psi_{n=2} | \psi_{n=1} \rangle \)
    4. \(\langle \psi_{n=2} | \psi_{n=2} \rangle \)

    \[ |\psi_1 \rangle = (\frac{2}{L})^{\frac{1}{2}}sin\frac{\pi x}{L}\]

    \[ |\psi_2 \rangle = (\frac{2}{L})^{\frac{1}{2}}sin\frac{2\pi x}{L}\]

    \[\langle \psi_1 |\psi_1 \rangle = \int \limits _{-\infty}^{\infty} \frac{2}{L}(sin\frac{\pi x}{L})^2 dx = 1\]

    \[\langle \psi_2 |\psi_1 \rangle = \int \limits _{-\infty}^{\infty} \frac{2}{L}(sin\frac{2\pi x}{L})(sin\frac{\pi x}{L}) dx = 0\]

    \[\langle \psi_1 |\psi_2 \rangle = \int \limits _{-\infty}^{\infty} \frac{2}{L}(sin\frac{\pi x}{L})(sin\frac{2\pi x}{L}) dx = 0\]

    \[\langle \psi_2 |\psi_2 \rangle = \int \limits _{-\infty}^{\infty} \frac{2}{L}(sin\frac{2\pi x}{L})^2 dx = 1\]

    Q7.9

    The wavefunction can be expanded into the complete set of basis of eigenstates of the Hamiltonian:

    \[| \Psi \rangle =\sum_i c_i | \phi_i \rangle \]

    What is the general expression of the off diagonal (\(i \neq j\)) and diagonal (\(i = j\)) matrix elements for the Hamiltonian in the basis set of its eigenstates?

    \[H_{ij} = \langle \phi_i | \hat{H} | \phi_j \rangle \]

    (hint: Apply the Hamiltonian \(\hat{H}\) on this arbitrary wavefuction \(| \Psi \rangle\) and its bra version \(\langle \Psi | \)).

    \[H_{ij} = \langle \phi_i |c_i \hat{H} c_j | \phi_j \rangle = (c_i \times c_j)E_j \langle \phi_i | \phi_j \rangle\ = 0\] Because \(\phi_i\) and \(\phi_j\) are orthogonal.

    \[H_{ii} = \langle \phi_i |c_i \hat{H} c_i | \phi_i \rangle = |c_i^2|E_i \langle \phi_i | \phi_i \rangle\ = E_i\] Because \(\phi_i\) is normalized.

    Q7.10 - Q7.12 moved to Homework 9

    Q7.13 (VERY optional)

    \[E_{trial}= \dfrac{\langle\psi_{trial}| H | \psi_{trial}\rangle}{\langle\psi_{trial} | \psi_{trial} \rangle}\]

    \[H = \dfrac{-\hbar^2}{2m} \dfrac{\delta^2}{\delta x^2}\]

    \[\langle\psi_{trial}|\psi_{trial}\rangle = \int_{0}^{L} N^2 e^{-2\beta x^2} dx = \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2} N^2 (2\beta)^{1/2}e^{-t^2}dt = 1\]

    where

    • \(t = (2\beta)^{1/2} x\) and
    • \(dt = (\beta)^{1/2} dx\)

    \[N = (\beta \pi)^{-1/4} \times (erf((2\beta)^{1/2} L))^{-1/2}\]

    Where \(erf((2\beta)^{1/2}L\) is the error function evaluated at \((2\beta)^{1/2}L\).

    \[ E_{trial} = {\langle\psi_{trial}| H | \psi_{trial}\rangle} = \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2} N^2 e^{-\beta x^2} \dfrac{-\hbar}{2m} \dfrac{\delta^2}{\delta x^2} e^{-\beta x^2} dx = \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2} N^2 e^{-\beta x^2} \dfrac{-\hbar}{2m} \dfrac{\delta}{\delta x} 2x\beta e^{-\beta x^2} dx\]

    \[ = \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2} N^2 \beta^{-1/2}e^{-t^2} \dfrac{-\hbar}{2m} (-2\beta e^{-t^2} + 4t^2\beta e^{-t^2})dt\]

    \[= -\frac{N^2 \hbar^2}{2m\beta^{1/2}}\left[\left(-2\beta \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2}e^{-t^2}dt\right) + \left(2\beta \int_{-(2\beta)^{1/2}L/2}^{(2\beta)^{1/2}L/2}t^2e^-t^2 dt\right)\right]\]

    \[= -\frac{N^2 \hbar^2 \beta^{1/2}}{2m}(\frac{L(2\beta)^{1/2}}{2e^{L^22\beta}} - \frac{1}{2}\pi^{1/2}ierfi(\frac{Li(2\beta)^{1/2}}{2}) -(2\beta \pi)^{1/2}erf(\frac{(2\beta)^{1/2}L}{2})))\]

    Where erfi is the imaginary error function. So overall we have for the unoptimized ground state energy:

    \[E_{trial} = -\frac{(\beta \pi)^{-1/2} \times (erf((2\beta)^{1/2} \frac{L}{2}))^{-1} \hbar^2 \beta^{1/2}}{2m}(-\frac{L(2\beta)^{1/2}}{2e^{\frac{L^2\beta}{2}}} + \frac{1}{2}\pi^{1/2}erf(\frac{-L(2\beta)^{1/2}}{2}) - (2\beta \pi)^{1/2}erf(\frac{(2\beta)^{1/2}L}{2}))\]

    Normally to optimize the energy we would take the derivative of our trial energy with respected to beta and find the minimum. Here due to the complexity of the equation an easier solution would be to numerically plot the energy as a function of beta, though to do that we would need to have a value for L.

    A) and B) A Gaussian is a poor trial function for a particle in a box since a Gaussian does not go to zero at any values other than infinity. While the box potential has a finite size, if the box potential had infinite size then we would be dealing with a free particle.

    C) This problem can be solved up to the unoptimized trial energy analytically but taking the derivative leads to a transcendental equation so it has to be solved numerically.


    #7A 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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