Question

*1.9 (b) For what potential energy function \(V(x)\) does \(\Psi\) satisfy the Schrödinger equation?

Answer 1

\[\Psi(x,t) = Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]
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(a) \[1=\int\limits_{-\infty}^{\infty}|\Psi|^2~\text d x\]\[1=\int\limits_{-\infty}^{\infty}\Psi^*\Psi~\text d x\]\[1=\int\limits_{-\infty}^{\infty}Ae^{ -a\left[\frac{mx^{2}}{\hbar}-it\right]}\cdot Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}~\text d x\]\[1=\int\limits_{-\infty}^{\infty}A^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]\[\text{let}\quad \nu=2a\left[\frac{m}{\hbar}\right]\]\[x=u+a\]\[\text{ d} x = \text {d}u\]
\[\frac{1}{A^2}=\int\limits_{-\infty}^{\infty}e^{-\nu x^2}~\text d x\]\[\frac{1}{A^2}=\sqrt\frac{\pi}{\nu}\]\[A=\left(\frac{\nu}{\pi}\right)^\frac{1}{4}\]\[A=\left(\frac{2a[\frac{m}{\hbar}]}{\pi}\right)^\frac{1}{4}\]\[A=\sqrt[4]{\frac{2am}{\pi\hbar}}\]

\[\Psi(x,t) = \sqrt[4]{\frac{2am}{\pi\hbar}}e^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]

Answer 2

(b) \(V(x)=\quad\color\gray?\)

Answer 3

The Schrödinger equation\[i\hbar\frac{\partial \Psi}{\partial t }=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi\]

Answer 4

so i just need to take some partials

Answer 5

\[\frac{\partial\Psi}{\partial t} =\]
\[\frac{\partial^2\Psi}{\partial x^2} = \]

Answer 6

\[\Psi(x,t) = Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]
\[\Psi_t = -iaAe^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]
\[\Psi_x = \frac{-2amx}{\hbar}Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]

Answer 7

product rule i guess?

Answer 8

\[\Psi_{xx} = \frac{-2am}{\hbar}Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}+ \frac{4a^2m^2}{\hbar^2}Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]

Answer 9

have i made any mistakes in (b) so far?

Answer 10

*

Answer 11

\[\Psi(x,t) = Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}\]
\[\Psi_t = -iaAe^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}=-ia\Psi\]
\[\Psi_x = \frac{-2amx}{\hbar}Ae^{ -a\left[\frac{mx^{2}}{\hbar}+it\right]}=\frac{-2amx}{\hbar}\Psi\]
\[\Psi_{xx} = \frac{-2am}{\hbar}\Psi+ \frac{4a^2m^2}{\hbar^2}\Psi\]

Answer 12

\[i\hbar\frac{\partial \Psi}{\partial t }=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi\]
\[\Downarrow\qquad\quad\qquad\qquad\qquad\Downarrow\]
\[i\hbar(-ia\Psi)=-\frac{\hbar^2}{2m}\left(\frac{-2am}{\hbar}\Psi+ \frac{4a^2m^2x^2}{\hbar^2}\Psi\right)+V\Psi\]
\[a\hbar\Psi=a\hbar\Psi-{2a^2mx^2}\Psi+V\Psi\]
\[2a^2mx^2\Psi=V\Psi\]
\[V(x)=2a^2mx^2\]

Answer 13

I don't see any mistakes on part (b).
How do we do part (c) now?

Answer 14

ok

Answer 15

(c)
\[\langle x \rangle=\int\limits_{-\infty}^{\infty}x|\Psi|^2~\text d x\]

\[=\int\limits_{-\infty}^{\infty}xA^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]

\[=A^2\int\limits_{-\infty}^{\infty}xe^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]
\[=A^2\int\limits_{-\infty}^{\infty}xe^{ -\nu x^2}~\text d x\]
\[=0 \qquad\text{(odd function)}\]

Answer 16

\[\langle x^2 \rangle =\int\limits_{-\infty}^{\infty}x^2|\Psi|^2~\text d x\]\[=\int\limits_{-\infty}^{\infty}x^2A^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]\[=2A^2\int\limits_{0}^{\infty}x^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]
\(\text{let} \)
\({\nu=2a[\frac{m}{\hbar}]}\)
\(x=u+a\)
\({ {\text{ d} x = \text {d}u}}\)
\[=A^2\int\limits_{-\infty}^{\infty}x^2e^{ -\nu x^2}~\text d x\]

Answer 17

now integration by parts , i suppose

Answer 18

Yes,integration by parts.
x^2e^-(vx)=x*xe^-(vx)
You should integrate xe^(-vx) and take the derivative of x.
To integrate xe^(-vx):
z substitution :
z=-vx
dz=-v

Answer 19

\[=A^2\int\limits_{-\infty}^{\infty}x^2e^{ -\nu x^2}~\text d x\]
\(∫p dq=pq-∫q dp\)

\(p=x^2\qquad \text dq= e^{\nu x^2}\)
\(dp=2x\text dx\qquad q= \frac{e^{\nu x^2}}{-2x\nu}\)

\[=A^2\left[\left.x^2\frac{e^{\nu x^2}}{-2x\nu}\right|-\int\frac{e^{\nu x^2}}{-2x\nu}2x\text dx\right]\]
\[=A^2\left[\left.x\frac{e^{\nu x^2}}{-2\nu}\right|_{-\infty}^{\infty}+\frac{1}{\nu}\int\limits_{-\infty}^{\infty}{e^{\nu x^2}}\text dx\right]\]

Answer 20

You have the ground-state wavefunction of a SHO, so when you're done you can check your results against the known results for the 1-D SHO.