Paul Lea Homework 2

Problem 9

$$-\frac{\hslash }{2m}\frac{{\partial }^2(A\cos (kx-\omega t))}{\partial x^2}=i\hslash \frac{\partial (A\cos (kx-\omega t))}{\partial t}$$ $$-\frac{\hslash }{2m}(-A{k}^2\cos (kx-\omega t))=i\hslash (A\omega \sin (kx-\omega t))$$ $$\frac{k^2\cos (kx-\omega t)}{2m}\ne i\omega \sin (kx-\omega t)$$

Therefore this is not a solution to the Schrodinger equation for a free particle

$$-\frac{\hslash }{2m}\frac{{\partial }^2(A\sin (kx-\omega t))}{\partial x^2}=i\hslash \frac{\partial (A\sin (kx-\omega t))}{\partial t}$$ $$-\frac{\hslash }{2m}(-A{k}^2\sin (kx-\omega t))=i\hslash (-A\omega \cos (kx-\omega t))$$ $$\frac{k^2\sin (kx-\omega t)}{2m}\ne -i\omega \cos (kx-\omega t)$$

Therefore this is not a solution to the Schrodinger equation for a free particle

Problem 10

$$-\frac{{\hslash }^2}{2m}\frac{\partial (c_1{\psi }_1+c_2{\psi }_2)}{\partial x}=i\hslash \frac{\partial (c_1{\psi }_1+c_2{\psi }_2)}{\partial t}$$ $$-\frac{{\hslash }^2}{2m}\left(\frac{\partial c_1{\psi }_1}{\partial x}+\frac{\partial c_2{\psi }_2}{\partial x}\right)=i\hslash (\frac{\partial c_1{\psi }_1}{\partial t}+\frac{\partial c_2{\psi }_2}{\partial t})$$ $$c_1{\psi }_1\propto {\psi }_1;c_2{\psi }_2\propto {\psi }_2$$ $$-\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_1}{\partial x}-\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_2}{\partial x}=i\hslash \frac{\partial {\psi }_1}{\partial t}+i\hslash \frac{\partial {\psi }_2}{\partial t}$$ $$-\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_1}{\partial x}=i\hslash \frac{\partial {\psi }_1}{\partial t}$$ $$-\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_2}{\partial x}=i\hslash \frac{\partial {\psi }_2}{\partial t}$$ $$\therefore -\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_1}{\partial x}+\left(-\frac{{\hslash }^2}{2m}\frac{\partial {\psi }_2}{\partial x}\right)=i\hslash \frac{\partial {\psi }_1}{\partial t}+i\hslash \frac{\partial {\psi }_2}{\partial t}$$

Problem 11

$$((A{e}^{-ikx}+B{e}^{ikx})(Aik{e}^{ikx}-Bik{e}^{-ikx})-(A{e}^{ikx}+B{e}^{-ikx})(-Aik{e}^{-ikx}+Bik{e}^{ikx}))$$

Expanded and simplified with wolfram

$$2ik(A^2-B^2)$$ $$j_x=\frac{\hslash }{2mi}(2ik(A^2-B^2))$$ $$j_x=\frac{\hslash }{m}k(A^2-B^2)$$

Problem 12

$$1=\underset{0}{\overset{L}{\int }}N{x}^2(L-x)dx$$ $$\frac{1}{N}=\underset{0}{\overset{L}{\int }}{x}^2(L-x)dx$$

Computed with wolfram

$$\frac{1}{N}=\frac{L^4}{12}$$ $$N=\frac{12}{L^4}$$

Finding expectation value for x

$$⟨x⟩=\int {\psi }^{\ast }x\psi dx$$ $$\underset{0}{\overset{L}{\int }}(\frac{12}{L^4}{x}^2(L-x){)}^{2}xdx$$ $$\frac{12}{L^4}\underset{0}{\overset{L}{\int }}(x^2(L-x^2){)}^{2}xdx$$

Computed with wolfram

$$\frac{12}{L^4}(\frac{L^8\cdot \left(6L^2-15L+10\right)}{60})$$ $$⟨x⟩=\frac{L^4\cdot \left(6L^2-15L+10\right)}{5}$$

Problem 13

$$⟨p_x⟩=\underset{-\infty }{\overset{\infty }{\int }}{\Psi }^{\ast }\frac{\hslash }{i}\frac{\partial }{\partial x}(\Psi )dx$$ $$\underset{-\infty }{\overset{\infty }{\int }}{e}^{-ikx}{\psi }^{\ast }(x)\frac{\hslash }{i}ik{e}^{ikx}\psi (x)dx$$ $$\underset{-\infty }{\overset{\infty }{\int }}{e}^{-ikx}\psi (x)\frac{\hslash }{i}ik{e}^{ikx}\psi (x)dx$$ $$\hslash k\underset{-\infty }{\overset{\infty }{\int }}{\psi }^2(x)dx$$ $$\underset{-\infty }{\overset{\infty }{\int }}{\psi }^2(x)dx=1$$ $$⟨p_x⟩=\hslash k$$

Problem 14

$$A(k)={\left(\frac{2}{\pi }\right)}^{1/4}\sqrt{\sigma }{e}^{-{\sigma }^2}(k-k_0{)}^2$$ $$\Psi (x)=\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}A(k)e^{ikx}dk$$ $$\Psi (x)=\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{\left(\frac{2}{\pi }\right)}^{1/4}\sqrt{\sigma }{e}^{-{\sigma }^2(k-k_0{)}^2}{e}^{ikx}dk$$ $$\underset{-\infty }{\overset{\infty }{\int }}{e}^{-a(k-k_0{)}^{2}x}{e}^{ikx}dx$$

Multiply by

$$e^{i{k}_{0}x}{e}^{-i{k}_{0}x}$$

You get

$$e^{i{k}_{0}x}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-a(k-k_0{)}^2}{e}^{ikx}{e}^{-i{k}_{0}x}dk$$ $$e^{i{k}_{0}x}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-a(k-k_0{)}^2}{e}^{i(k-k_0)x}dk$$ $$e^{i{k}_{0}x}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{k}^{\prime 2}}{e}^{i{k}^{\prime }x}d{k}^{\prime }$$ $$e^{i{k}_{0}x}\underset{-\infty }{\overset{\infty }{\int }}{e}^{-a{k}^{\prime 2}+i{k}^{\prime }x}d{k}^{\prime }$$ $$(b{k}^{\prime }+c{)}^2=b^2{k}^{\prime 2}+2cb{k}^{\prime }+c^2$$ $$b^2{k}^{\prime 2}+2cb{k}^{\prime }=(b{k}^{\prime }+c{)}^2-c^2$$ $$a{k}^{\prime 2}+i{k}^{\prime }x$$ $$b^2=a$$ $$a{k}^{\prime 2}+i{k}^{\prime }x=\left(\sqrt{a{k}^{\prime }}+\frac{ix}{2\sqrt{a}}\right)--\frac{x^2}{4a}$$ $$2bc=ix$$ $$c=\frac{ix}{2\sqrt{a}}$$

rewrite int as

$$e^{i{k}_{0}x}\int e^{-(\sqrt{a}{k}^{\prime }+(ix)/2\sqrt{a}{)}^2}{e}^{-x^2/4a}dx$$ $$e^{i{k}_{0}x}\int e^{-u^2}{e}^{-x^2/4a}dx$$ $$e^{i{k}_{0}x}{e}^{-x^2/4\sigma }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-u^2}du=\sqrt{\pi }$$ $$u=-(\sqrt{a}{k}^{\prime }+(ix)/2\sqrt{a}{)}^2;du=\frac{a}{2}(x-2ik)$$

We can drop the constants

$$\propto e^{i{k}_{0}x}{e}^{-x^2/4\sigma }$$

Ta-da

Problem 15 Normalize Wave functions

$$\underset{-\infty }{\overset{\infty }{\int }}{\psi }^{\ast }\psi dx=1$$ $$1=\underset{-\infty }{\overset{0}{\int }}{A}^2{e}^{2\kappa x}dx+\underset{0}{\overset{\infty }{\int }}{A}^2{e}^{-2\kappa x}dx$$ $$\frac{1}{A^2}=\underset{-\infty }{\overset{0}{\int }}{e}^{2\kappa x}dx+\underset{0}{\overset{\infty }{\int }}{e}^{-2\kappa x}dx$$ $$\frac{1}{A^2}=\frac{1}{2\kappa }+\frac{1}{2\kappa }$$ $$\frac{1}{A^2}=\frac{1}{\kappa }$$ $$A^2=\kappa \therefore A=\sqrt{\kappa }$$

Finding $⟨x⟩$

$$\underset{-\infty }{\overset{0}{\int }}x\kappa e^{2\kappa x}dx+\underset{0}{\overset{\infty }{\int }}x\kappa e^{-2\kappa x}dx$$

Integration by parts (wolfram)

$$-\frac{1}{4\kappa }+\frac{1}{4\kappa }$$ $$⟨x⟩=0$$

This corresponds to the graphical observation!

$$\underset{-\infty }{\overset{0}{\int }}{x}^2\kappa e^{2\kappa x}dx+\underset{0}{\overset{\infty }{\int }}{x}^2\kappa e^{-2\kappa x}dx$$

Use integration by parts (wolfram alpha)

$$\frac{1}{4{\kappa }^2}+\frac{1}{4{\kappa }^2}$$ $$\frac{2}{4{\kappa }^2}=\frac{1}{2{\kappa }^2}$$ $$\frac{1}{2{\kappa }^2}$$ $$\Delta x=\sqrt{\frac{1}{2{\kappa }^2}+0^2}=0$$ $$\Delta x=\frac{1}{\kappa }\sqrt{\frac{1}{2}}$$