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Show that the function satisfies the Schrodinger eqaution

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data

    So im 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, im im not sure how to begin with these types of questions since they are new to me.

    Quantum mechanics Qns.PNG
    2. Relevant equations

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

    3. The attempt at a solution

    I dont know where to start for the first, i could try differentiating it, but im 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 im guessing would be of the form GMm/r^2 in the unit x direction...

    Thanks for any pointers!
     
  2. jcsd
  3. Dec 4, 2014 #2

    ShayanJ

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    Gold Member

    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 [itex] V=-\int_{r_{ref}}^r \vec g \cdot \vec{dr} [/itex].
     
  4. Dec 6, 2014 #3
    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 cant 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?
     
  5. Dec 6, 2014 #4

    ShayanJ

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    Your wave-function is the superposition of plane waves of all possible wave-numvers(Lets call them modes). Now for each mode([itex] \varphi_k(x) [/itex]), we know that its energy is only kinetic and so we have [itex] \hat H \varphi_k(x)=\frac{\hbar^2 k^2}{2m}\varphi_k(x) [/itex]. We also know that [itex] i\hbar \partial_t \varphi_k(x)=\hbar \omega \varphi_k(x) [/itex]. So from Schrodinger equation, we have [itex] \hbar \omega=\frac{\hbar^2 k^2}{2m} [/itex].
    I told you about part two, but looks like you didn't pay attention. [itex] 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)[/itex], where [itex] x_0 [/itex] is the origin of the potential which is arbitrary.
     
  6. Dec 6, 2014 #5
    I am not familiar with the notation ##\hat H##, does it represent the total energy of the system? I have not learnt about Hamiltonians yet, would you reccomend that i do so before I learn QM? I think the book im 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.
     
  7. Dec 6, 2014 #6

    ShayanJ

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