2.4: Problems
- Page ID
- 106809
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Note: Always express angles in radians (e.g. \(\pi/2\), not \(90^{\circ}\)). When expressing complex numbers in Cartesian form always finish your work until you can express them as \(a+bi\). For example, if you obtain \(\frac{2}{1+i}\), multiply and divide the denominator by its complex conjugate to obtain \(1-i\).
Remember: No calculators allowed!
Given \(z_1=1+i\), \(z_2=1-i\) and \(z_3=3e^{i \pi/2}\), obtain:
- \(z_1 z_2\)
- \(z_1^2\)
- \(2z_1-3z_2\)
- \(|z_1|\)
- \(2z_1-3z_2^*\)
- \(\frac{z_1}{z_2}\)
- Express \(z_2\) as a complex exponential
- \(|z_3|\)
- \(z_1+z_3\), and express the result in cartesian form
- Display the three numbers in the same plot (real part in the \(x\)-axis and imaginary part in the \(y\)-axis)
The following family of functions are encountered in quantum mechanics:
\[\Phi_m(\phi)=\frac{1}{\sqrt{2 \pi}}e^{i m \phi}, m= 0, \pm 1,\pm 2, \pm 3 \dots, 0 \le \phi \le 2\pi \nonumber \]
Notice the difference between \(\Phi\) (the name of the function), and \(\phi\) (the independent variable). The definition above defines a family of functions (one function for each value of \(m\)). For example, for \(m=2\):
\[\Phi_2(\phi)=\frac{1}{\sqrt{2 \pi}}e^{2i \phi}, \nonumber \]
and for \(m=-2\):
\[\Phi_{-2}(\phi)=\frac{1}{\sqrt{2 \pi}}e^{-2i \phi}, \nonumber \]
- Obtain \(|\Phi_m(\phi)|^2\)
- Calculate \(\int_0 ^{2\pi}|\Phi_m(\phi)|^2 \mathrm{d}\phi\)
- Calculate \(\int_0 ^{2\pi}\Phi_m(\phi)\Phi_n^*(\phi) \mathrm{d}\phi\) for \(m \neq n\)
- Calculate \(\int_0 ^{2\pi}\Phi_m(\phi) \mathrm{d}\phi\) for \(m = 0\)
- Calculate \(\int_0 ^{2\pi}\Phi_m(\phi) \mathrm{d}\phi\) for \(m \neq 0\)
Given the function
\[f(r,\theta,\phi)=4 r e^{-2r/3} \sin{\theta}e^{-2i\phi/5} \nonumber \]
Write down an expression for \(|f(r,\theta,\phi)|^2\)