11.4: Problems
- Page ID
- 106872
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider the operator \(\hat A\) defined in Equation \(11.1.1\) as \(\hat A=\hat x + \dfrac{d}{dx}\). Is it linear or non-linear? Justify.
Which of these functions are eigenfunctions of the operator \(-\frac{d^2}{dx^2}\)? Give the corresponding eigenvalue when appropriate. In each case \(k\) can be regarded as a constant.
\[f_1(x)=e^{ikx} \nonumber \]
\[f_2(x)=\cos(kx) \nonumber \]
\[f_3(x)=e^{-kx^2} \nonumber \]
\[f_4(x)=e^{ikx}-cos(kx) \nonumber \]
In quantum mechanics, the \(x\), \(y\) and \(z\) components of the angular momentum are represented by the following operators:
\[ \begin{align*} \hat{L}_x &=i\hbar\left(\sin\phi\frac{\partial}{\partial \theta}+\frac{\cos\phi}{\tan \theta}\frac{\partial}{\partial\phi}\right) \\[4pt] \hat{L}_y &=i\hbar\left(-\cos\phi\frac{\partial}{\partial \theta}+\frac{\sin\phi}{\tan \theta}\frac{\partial}{\partial\phi}\right) \\[4pt] \hat{L}_z &=-i\hbar\left(\frac{\partial}{\partial \phi}\right) \end{align*} \]
The operator for the square of the magnitude of the orbital angular momentum, \(\hat{L}^2=\hat{L}^2_x +\hat{L}^2_y+\hat{L}^2_z\) is:
\[\hat{L}^2=-\hbar^2\left(\frac{\partial^2}{\partial \theta^2}+\frac{1}{\tan \theta}\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right) \nonumber \]
a) Show that the three 2p orbitals of the H atom are eigenfunctions of both \(\hat{L}^2\) and \(\hat{L}_z\), and determine the corresponding eigenvalues.
\[\psi_{2p0}=\frac{1}{\sqrt{32\pi a_0^3}}r e^{-r/2 a_0}\cos\theta \nonumber \]
\[\psi_{2p+1}=\frac{1}{\sqrt{64\pi a_0^3}}r e^{-r/2 a_0}\sin\theta e^{i\phi} \nonumber \]
\[\psi_{2p-1}=\frac{1}{\sqrt{64\pi a_0^3}}r e^{-r/2 a_0}\sin\theta e^{-i\phi} \nonumber \]
b) Calculate \(\hat{L}_x\psi_{2p0}\). Is \(\psi_{2p0}\) and eigenfunction of \(\hat{L}_x\)?
c) Calculate \(\hat{L}_y\psi_{2p0}\). Is \(\psi_{2p0}\) and eigenfunction of \(\hat{L}_y\)?
Prove that
\[\left[\hat{L}_z,\hat{L}_x\right]=i\hbar \hat{L}_y \nonumber \]
For a system moving in one dimension, the momentum operator can be written as
\[\hat p = i \hbar \frac{d}{dx} \nonumber \]
Find the commutator \([\hat x, \hat p]\)
Note: \(\hbar\) is defined as \(h/{2 \pi}\), where \(h\) is Plank’s constant. It has been defined because the ratio \(h/{2 \pi}\) appears often in quantum mechanics.
We demonstrated that \(\psi_1s\) is not an eigenfunction of \(\hat T\). Yet, we can calculate the average kinetic energy of a 1s electron, \(\left \langle T \right \rangle\). Use Equation \(11.3.1\) to calculate an expression for \(\left \langle T \right \rangle\).
Use the Hamiltonian of Equation \(11.3.5\) to calculate the energy of the electron in the 1s orbital of the hydrogen atom. The normalized wave function of the 1s orbital is:
\[\psi=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0} \nonumber \]
The expression of Equation \(11.3.1\) can be used to obtain the expectation (or average) value of the observable represented by the operator \(\hat{A}\).
The state of a particle confined in a one-dimensional box of length a is described by the following wavefunction:
\[\psi(x)=\begin{cases} \sqrt{\frac{2}{a}}\sin\left(\frac{\pi x}{a} \right )& \mbox{ if } 0\leq x\leq a \\ 0 &\mbox{otherwise} \end{cases} \nonumber \]
The momentum operator for a one-dimensional system was introduced in Problem \(\PageIndex{5}\).
a) Obtain an expression for \(\hat{p}^2\) and determine if \(\psi\) is an eigenfunction of \(\hat{p}\) and \(\hat{p}^2\). If possible, obtain the corresponding eigenvalues.
Hint: \(\hat{p}^2\) is the product \(\hat{p}\hat{p}\).
b) Determine if \(\psi\) is an eigenfuction of \(\hat{x}\). If possible, obtain the corresponding eigenvalues.
c) Calculate the following expectation values: \(\left \langle x \right \rangle\), \(\left \langle p^2 \right \rangle\), and \(\left \langle p \right \rangle\). Compare with the eigenvalues calculated in the previous questions.