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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]

    become

    [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

    Mark44

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

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

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

    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

    Mark44

    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,
    WHY DONT TEXT BOOKS JUST SHOW HOW THINGS ARE DONE STEP BY STEP?
    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

    Mark44

    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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