Question

Differential Eq.

Answer 1

\[ih \frac{ d }{ dt }\psi+\frac{ h^2 }{ 2m }\frac{ d^2 }{ dx^2 }\psi\]

Answer 2

\[\psi(x=0,t)=0, \psi(x=L,t)=0\]

Answer 3

@Loser66

Answer 4

ohoh... I don't know.

Answer 5

PDE?

Answer 6

yup

Answer 7

=0, right?
If I replace \(\psi\) by \(u\), I can rewrite it as \(ih*U_t +\dfrac{h^2}{2m}U_{xx}=0\)

Answer 8

Solution\[\psi(x,t)= \frac{ 4 }{ \pi \sqrt{L} } \frac{ 1 }{ n } \sin(kx) e^{-i \frac{ E }{ h }t}\]

Answer 9

ok, What do you want me to do? find solution or verify the solution?

Answer 10

I got the solution, but I can't figure out what the question wants! It says take the time derivative of P and use SE to get dpsi/dt

Answer 11

The solution is given, right?

Answer 12

yup

Answer 13

If we have to verify, then, it is harder than find it out. ok, I need time to work on it. But not promise that I will have it done. hehehe..

Answer 14

Lol, this is probably beyond me, but I'll try and help as far as I can...

Answer 15

Firstly, you have calculated P right?

Answer 16

\[P= \frac{ 16 }{ \pi^2 L} \frac{ 1 }{ n^2} \sin(kx) \sin(kx)\]

Answer 17

O.o oh, so any variables with respect to time disappears? O.o

Answer 18

okay, then take the partial derivative of P wrt t

Answer 19

wouldn't it be same as ^

Answer 20

Hmmm, I believe you differentiate with P wrt t when P is equal to your thing above. But then you also differentiate \(\psi^*\psi\)

Answer 21

I am pretty sure this is the Schrödinger equation for a particle in infinite potential well. No idea what the question is asking though.

Answer 22

this, probably: https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Local_conservation_of_probability :P

Answer 23

@Kainui

Answer 24

solution I got \[\psi(x,t)= \sum_{}^{n}C_n \sin(k_nx)e^{\frac{ -iEt}{ h}}\]

Answer 25

\[k_n= \frac{ n*\pi }{ L}\]

Answer 26

They want you to "plug in the Schrodinger equation" when you integrate the probability. What they mean is, the SE looks like this:

\[\dot \psi = -i \hbar \hat H \psi\]
so that is also, literally the time derivative of \(\psi\).

Follow the problem step by step, don't try to "make stuff up". Start here:

"To demonstrate this consider the local probability density \(P(x,t)=\psi^* \psi\)." Then in the very next sentence they tell you exactly what you need to do. "Take the time derivative of P and use the Schrodinger Equation to evaluate \(\dot \psi\)." Do this.

They proceed to give you more details after that of how you do this, which is pretty much what I said earlier. Follow this, show me your steps, and if you get stuck/confused we'll deal with it when it pops up.

Answer 27

\[dp/dt= \psi*' \psi'\]

Answer 28

\[\psi*'=ih H \psi\]

Answer 29

\[\int\limits_{0}^{L}p'=-ihH \psi(ih H \psi) dt\]

Answer 30

\[\int\limits_{0}^{L}p'=\int\limits_{0}^{L}\frac{ h^2 }{ 4m^2}\frac{ d^2 \psi }{ dx^2 }\frac{ d^2 \psi }{ dx^2 }dt\]

Answer 31

Wait a sec, this isn't the product rule!!!
\[P' \ne {\psi^*}' \psi'\]
Here we go:
\[P' = {\psi^*}' \psi+ {\psi^*} \psi'\]

Answer 32

Okay

Answer 33

\[\psi'=\frac{ ih }{ 2m}\frac{ d^2\psi }{ dx^2 }\]

Answer 34

then what is psi?

Answer 35

the complex conjugate also stays on the derivative of psi in the second equation you gave too.

psi is the wave function

Answer 36

\[\psi'^*=\frac{ ih }{ 2m}\frac{ d^2\psi^* }{ dx^2 }\]

Answer 37

is psi solution that wrote ^