Question

Calculate the expectation value of x squared

\[\langle x^2\rangle \]

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

Answer 1

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

Answer 2

\[=\int\limits_{-\infty}^{\infty}x^2A^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]
\[=2\sqrt{\frac{2am}{\pi\hbar}}\int\limits_{0}^{\infty}x^2e^{ -2a\left[\frac{mx^{2}}{\hbar}\right]}~\text d x\]
\[\text{let } \nu=\frac{2am}{\hbar}\]
\[=2\sqrt{\frac{\nu}{\pi}}\int\limits_{0}^{\infty}x^2e^{ -\nu x^2}~\text d x\]

Answer 3

\(\int p \text dq=\left.pq\right|-\int q \text dp\)


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

\[=2\sqrt{\frac{\nu}{\pi}}\left[\left.x^2\frac{e^{-\nu x^2}}{-2x\nu}\right|_{0}^{\infty}-\int\limits_{0}^{\infty}\frac{e^{-\nu x^2}}{-2x\nu}2x\text dx\right]\]

\[=2\sqrt{\frac{\nu}{\pi}}\left[\left.x\frac{e^{-\nu x^2}}{-2\nu}\right|_{0}^{\infty}+\frac{1}{\nu}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\right]\]

Answer 4

\[=\frac{2}{\nu}\sqrt{\frac{\nu}{\pi}}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\]

\[=\frac{2}{-2\nu^2 x}\sqrt{\frac{\nu}{\pi}}\left.{e^{-\nu x^2}}\right|_{0}^{\infty}\]


\[=-\frac{1}{ x}\frac{\nu^{-3/2}}{\sqrt{\pi}}\left.{e^{-\nu x^2}}\right|_{0}^{\infty}\]

Answer 5

?

Answer 6

your p should be x and
dq x exp(stuff) dx

Answer 7

i thought\[\int\limits_{0}^{\infty}e ^{-x ^{2}}d x=\frac{ \sqrt{\pi} }{ 2 } \] ???

Answer 8

but why @phi?

good point @jeanlouie ,
what was i thinking

Answer 9

you integrated dq and got a bogus result

you want to integrate exp(u) du
u in this case is x^2 (ignore constants)
and du is 2 x dx
so you need an x dx to integrate exp(x^2)

Answer 10

\[=\frac{2}{\nu}\sqrt{\frac{\nu}{\pi}}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\]\[=\frac{2}{\nu}\sqrt{\frac{\nu}{\pi}}\sqrt{\frac{\pi}{4\nu}}\]\[=\frac 1 \nu\]\[=\frac1{{2am}/{\hbar}}\]\[=\frac{\hbar}{2am}\]

Answer 11

the answer in the back of my book says im off by a factor of two

Answer 12

I gotta agree with @phi You would need chain rule to take the derivative of q.

Answer 13

Sorry, not chain rule, quotient rule.

Answer 14

hmm

Answer 15

error function?

Answer 16

What if you tried this:

Answer 17

Kinda mess but:

Answer 18

\[=2\sqrt{\frac{\nu}{\pi}}\int\limits_{0}^{\infty}x^2e^{ -\nu x^2}~\text d x\]\(\int p \text dq=\left.pq\right|-\int q \text dp\)
\(p=x\qquad\text dq= xe^{-\nu x^2}\text dx\)
\(\text dp=\text dx\qquad q= \frac{e^{-\nu x^2}}{-2\nu}\)

\[=2\sqrt{\frac{\nu}{\pi}}\left[\left.x\frac{e^{-\nu x^2}}{-2\nu}\right|_{0}^{\infty}+\frac{1}{2\nu}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\right]\]\[=\frac{2{}\sqrt{\frac{\nu}{\pi}}}{2\nu}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\]\[=\frac{1}{\nu}\sqrt{\frac{\nu}{\pi}}\int\limits_{0}^{\infty}{e^{-\nu x^2}}\text dx\]\[=\frac{1}{\nu}\sqrt{\frac{\nu}{\pi}}\sqrt{\frac{\pi}{4\nu}}\]\[=\frac 1 {2\nu}\]\[=\frac1{2\times2a[\frac{m}{\hbar}]}\]\[=\frac{\hbar}{4am}\]\[\Large\color{red}\checkmark\]

Answer 19

Cool. Although I have to say I didn't understand a single bit of the first part of the problem!

Answer 20

thankyou to
phi, cruffo, jeanlouie

Answer 21

what bit didn't you understand @cruffo?

Answer 22

Well, I know what it means to find the expected value of a distribution. It's the first 3 lines of the discussion that I'm not sure what happened.

Answer 23

Sorry, not of "of a distribution" should have been "of a random variable".

Answer 24

oh i think i used some information from previous parts of the question

Answer 25

Ahhh..
Had to prove something about the Schrödinger equation in spherical coordinates in DE. What class is if for btw?

Answer 26

QM

Answer 27

figures. Good luck!

Answer 28

\[\langle p \rangle = \int\limits_{-\infty}^{\infty}\frac{\hbar}{i} \frac{\partial}{\partial x} |\Psi|^2~\text d x\]\[=\int\limits_{-\infty}^{\infty}\frac{\hbar}{i} \frac{\partial}{\partial x} \sqrt{\frac{\nu}{\pi}}e^{ -\nu x^2}\text dx\]\[= \frac{\hbar}{i} \sqrt{\frac{\nu}{\pi}}\int\limits_{-\infty}^{\infty}\frac{\partial}{\partial x}e^{ -\nu x^2}\text dx\]\[= \frac{\hbar}{i} \sqrt{\frac{\nu}{\pi}}\int\limits_{-\infty}^{\infty}-2\nu xe^{ -\nu x^2}\text dx\]\[= -2\nu \frac{\hbar}{i} \sqrt{\frac{\nu}{\pi}}\int\limits_{-\infty}^{\infty}xe^{ -\nu x^2}\text dx\]\[=0 \qquad\text{(odd function)}\]