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Homework Help: Quick integration question!

  1. Apr 19, 2010 #1
    Not really a homework question, but

    After separation of variables of the time independant schrodinger equation i get, one equation

    [tex]\frac{d \varphi}{\partial t} = -\frac{i}{\hbar}E \varphi dt[/tex]

    which i need to solve,
    so i multiply through by dt and integrate giving me

    [tex] \int d\varphi = \int -\frac{i}{\hbar} E \varphi dt [/tex]

    now I realize that some of these variables are constants, like [tex]\hbar[/tex]

    But I don't know how to integrate the right side of that equation
    so what I'm asking

    is how does,
    [tex] \int -\frac{i}{\hbar} E \varphi dt [/tex]


    [tex] e^{-iE\frac{t}{\hbar}} [/tex]
    could someone please explain it?

    (i'm not very good at this kind of integration,

    how do i know when integrating something, that it becomes an exponential function?
  2. jcsd
  3. Apr 19, 2010 #2
    thank you in advance.
  4. Apr 19, 2010 #3


    Staff: Mentor

    Is this your differential equation?
    [tex]\frac{d \phi}{dt} = -\frac{i}{\hbar}E \phi [/tex]

    In other words, it's just the derivative of phi with respect to t on the left side, and there's no dt on the right side.

    Any time you have a differential equation of the form dy/dt = Ky, the solution is an exponential function: y(t) = AeKt.

    The only function whose derivative is a constant multiple of itself is eKt.

  5. Apr 19, 2010 #4
    so because [tex] -\frac{i}{\hbar} E [/tex] is constant, (calling it K)

    [tex] \frac{dy}{dt} [/tex] is equivalent to my [tex] \frac{d\phi}{dt} [/tex]

    and Ky is my [tex] K \phi [/tex]

    giving [tex] \phi (t) = e^{-\frac{i}{h}Et} [/tex]

    I don't really know how it is a constant multiple of itself when differentiated

    [tex] Ae^{-\frac{i}{h}E} [/tex] differentiated dosen't look like [tex] -\frac{i}{h} E \phi [/tex] to me :(
  6. Apr 19, 2010 #5
    I don't know much about physics, but I'm going to assume that your first equation should be
    \frac{d \varphi}{dt} = -\frac{i}{\hbar}E \varphi

    Now you want to solve for [tex]\varphi[/tex], so get it alone and multiply through by dt:

    \frac{d \varphi}{\varphi} = -\frac{i}{\hbar}E dt

    Integrate both sides:

    [tex] \ln \varphi = -\frac{i}{\hbar}E t [/tex]

    Exponentiate both sides, and you are done.
  7. Apr 19, 2010 #6
    Thank you!

    I didn't see that sneaky [tex]\phi [/tex] movement to become [tex] ln \phi [/tex]

    i understand and will remember this rule forever :P
  8. Apr 19, 2010 #7


    Staff: Mentor

    [tex] \phi (t) = e^{-\frac{i}{\hbar}Et} [/tex]
    [tex] \Rightarrow \phi '(t) = \frac{-iE}{\hbar}e^{-\frac{i}{\hbar}Et} = \frac{-iE}{\hbar}\phi (t)[/tex]

  9. Apr 20, 2010 #8
    As you do more differential equations work, you will become accustomed to this very common situation where when you have dy/dx = Ay, you get y =ceAx. You can do the steps from my post (move the y to other side and multiply by dx), but it will become unnecessary, you'll just recognize this situation and automatically know the solution.
  10. Apr 20, 2010 #9
    Thanks everyone I really understand how that type of differential equation works now,
    damn you griffiths lol.

    wish all exemplar questions were structured like this: (also my finished differential eq)

    [tex] \int \frac{d\phi}{dt} = \int -\frac{i}{\hbar}E\phi [/tex]

    multiplying by dt

    [tex] \int \frac{d\phi}{dt}dt = \int -\frac{i}{\hbar}E\phi dt [/tex]

    canceling out the dt, and moving the [tex]\phi[/tex] to the left side of the equation

    [tex] \int \frac{d\phi}{\phi} = \int -\frac{i}{\hbar}Edt[/tex]

    the left hand side of the equation is the same as 1/phi ,

    [tex] \int \frac{1}{\phi}d\phi =\int -\frac{i}{\hbar}Edt[/tex]

    i,hbar,E are constants
    and integrating that I get

    [tex] ln\phi = -\frac{i}{\hbar}Et[/tex]

    sovling for phi i get,

    [tex] eln\phi = e-\frac{i}{\hbar}Et[/tex]

    canceling out I get
    [tex] \phi = e^{-\frac{i}{\hbar}Et[/tex]

    (note I know there's also I constant of integration, with the e^i/hbar et, BUT griffiths says it gets absorbed into the wave function in the schrodinger equation, and I understand how it works, that's why I left it out)

    thanks to all,
    i'm learning all these mathematical tricks that are making life so much easier,
    I've done 5 physics papers and 1 mathematics paper so far, I've been told that if you're not good at one you're not any good at the other, but I'm getting better

    cheers mark ;P
  11. Apr 20, 2010 #10


    Staff: Mentor

    Your first two steps should be collapsed into one. When you integrate you should specify the variable of integration, which is t in this case. So your first two steps should be:
    [tex] \frac{d\phi}{dt} = -\frac{i}{\hbar}E\phi [/tex]
    [tex] \Rightarrow \int \frac{d\phi}{dt}dt = \int -\frac{i}{\hbar}E\phi dt [/tex]
    Your comment below about the constant notwithstanding, you could do this:
    [tex] ln\phi = -\frac{i}{\hbar}Et + C[/tex]
    [tex] \Rightarrow e^{ln\phi} = e^{-\frac{i}{\hbar}Et + C} = e^{-\frac{i}{\hbar}Et * e^C = Ae^{-\frac{i}{\hbar}Et [/tex]

    where A = eC

    In the step above you are exponentiating each side; that is, you are making each side the exponent on e.
    [tex] e^{ln\phi} = e^{-\frac{i}{\hbar}Et}[/tex]
    I agree that you are getting better. I've been following your work for some time, and it has shown considerable improvement.
    Last edited: Apr 20, 2010
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