Show that the function satisfies the Schrodinger eqaution

[pondzo]

Homework Statement

So I am quite new to quantum mechanics and i am self teaching through a book called "Quantum mechanics for applied physics and engineering" by Albert Thomas Fromhold Jr. There are no solutions to the exercises, I am im not sure how to begin with these types of questions since they are new to me.

Homework Equations

1) Time dependent one-dimensional S.W.E
2) Time dependant one-dimensional S.W.E with potential function

The Attempt at a Solution

I don't know where to start for the first, i could try differentiating it, but I am not sure how, maybe the fundamental theorem of calculus ?

As for the second, I know i need to include a potential function for gravity which I am guessing would be of the form GMm/r^2 in the unit x direction...

Thanks for any pointers!

 

[ShayanJ]

For the first, you should use the Leibniz integral rule.
And for the second, there are 2 problems with what you suggest:
1) The inverse square law is for the force, not for the potential.
2) The potential should be an accordance with the force. The potential 1/r is for the inverse square force. The uniform field of the question has a linear potential. In fact the potential is derived from $V=-\int_{r_{ref}}^r \vec g \cdot \vec{dr}$.

 

[pondzo]

Hi Shyan, thank you for the reply.

I have tried to apply Leibniz's Integral rule as you suggested and have run into what seems a dead end;

$\psi_{xx}(x,t)=\int_{-\infty}^{\infty}-k^2A(k)e^{i(kx-wt)}dk$

$\psi_{t}(x,t)=\int_{-\infty}^{\infty}-i\omega A(k)e^{i(kx-wt)}dk$

When I sub both of these into the one dimensional time dependent Schrodinger equation,$\frac{-\hbar^2}{2m}\psi_{xx}=i\hbar\psi_{t}$, I get;

$\frac{\hbar}{2m}\int_{-\infty}^{\infty}-k^2A(k)e^{i(kx-wt)}dk=\int_{-\infty}^{\infty}\omega A(k)e^{i(kx-wt)}dk$

I do not know where to proceed from here, I can't figure out how to make the integrals on either side of the equality look like each other (so i can cancel them) since i can't take k or $\omega$ out of the integral. I tried to use that k=w/v where v is velocity but to no avail. Another pointer would be greatly appreciated!

For part two, would the potential function be; $V(x)=\frac{-GMm}{x}$ ? Also can i assume that it is a one dimensional case, I suspect i can't since the hint involved the unit vector in the x direction, which seems to imply 3 dimensional space?

 

[ShayanJ]

Your wave-function is the superposition of plane waves of all possible wave-numvers(Lets call them modes). Now for each mode($\varphi_k(x)$), we know that its energy is only kinetic and so we have $\hat H \varphi_k(x)=\frac{\hbar^2 k^2}{2m}\varphi_k(x)$. We also know that $i\hbar \partial_t \varphi_k(x)=\hbar \omega \varphi_k(x)$. So from Schrodinger equation, we have $\hbar \omega=\frac{\hbar^2 k^2}{2m}$.
I told you about part two, but looks like you didn't pay attention. $V(x)=-\int_{x_0}^x g \hat x \cdot \vec {dx'}=-\int_{x_0}^x g \hat x \cdot \hat x dx'=-\int_{x_0}^x g dx'=-g(x-x_0)$, where $x_0$ is the origin of the potential which is arbitrary.

 

[pondzo]

I am not familiar with the notation $\hat H$, does it represent the total energy of the system? I have not learned about Hamiltonians yet, would you reccomend that i do so before I learn QM? I think the book I am reading will introduce it soon though...
I did manage to get it to $\omega=\frac{\hbar k^2}{2m}$ but i couldn't see how this confirmed it is a valid solution. I'm obviously missing something quite important.

 

[ShayanJ]

You should know what is a Hamiltonian so study about it. Anyway, $\hat H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)$(for one particle in 1D). But because in the first exercise, V(x)=0, we have $\hat H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$.
Just take the integrals to the same side and factor $A(k) e^{i(kx-\omega t)} dk$.