Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    So lately I've been working through a book on quantum mechanics and I have a couple of questions about solving the 1D Schrodinger wave equation.

    1. When solving the equation for a rectangular potential barrier:

    I understand that \displaystyle \frac{\hbar^2 }{2m} \frac{d^2\Psi }{dx^2}+(E-V)\Psi =0
    and for the classically forbidden region E-V<0 and by inspection a suitable wavefunction would be \displaystyle \Psi =Ae^{kx} and then by differentiating to find the second derivative and substituting, simplifying, solving for k, etc.. you get to a general solution of
    \displaystyle \Psi =Ae^{\frac{\sqrt{2m(V-E)} }{\hbar} x}+Be^{-\frac{\sqrt{2m(V-E)} }{\hbar} x}

    but how would you then solve for the probability that the particle has tunnelled through the potential barrier from this?

    2. When solving the equation for a quantised harmonic oscillator:

    \displaystyle \frac{\hbar^2 }{2m} \frac{d^2\Psi }{dx^2}+(E-\frac{kx^2}{2} )\Psi =0

    and again by inspection a suitable wavefunction is \diplaystyle \Psi =Ae^{-\frac{1}{2} \alpha x^2} and I know how to solve for \alpha , E, A but I don't get why this particular wavefunction leads to the value of E that is the zero-point energy.

    Would someone please explain these to me? Thanks.
    Offline

    2
    ReputationRep:
    There's a few ways of approaching this, but it's quite a bit more Mathematically Heavy than for a rectangular problem.

    Step 1: De-dimensionalise
     \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}} + \frac{mx^2\omega^2}{2} \psi = E\psi

    We can change a few things in this problem to make it simpler; in this case we choose that  x = yl and then we substitute this into the equation; by de-dimensionalising it allows us to compare some quantities later which we wouldn't be able to do otherwise.

    Therefore, we get:

     \frac{-\hbar^{2}}{2ml^2} \frac{d^{2}\psi}{dy^{2}} + \frac{y^2 l^2 m \omega^2}{2} \psi = E\psi

    We multiply through by  \frac{-2ml^2}{\hbar^2}

    and then choose that  \frac{m^2 l^4 \omega^2}{\hbar^2} = 1
    which gives that  l = (\frac{\hbar}{m \omega})^{\frac{1}{2}}

    And choose that  \epsilon = \frac{-2 E m l^2}{\hbar^2}

    This simplifies our differential equation significantly; we get that

     \frac{d^2 \psi}{dy^2} - y^2 \psi = -\epsilon \psi

    Step 2: Look at the behaviour as y -> +/- infinity

    Outside of the potential, we need the wavefunction to die away, so we assume that  y^2 >> \epsilon and therefore we can say that
     \psi '' - y^2 \psi = 0 which has the solution
     \psi = e^{-\frac{1}{2}y^2}
    This accounts for the behaviour outside of the potential but not that inside; we say that  \Psi = \psi \phi_\epsilon and obviously we know psi. By substituting into the Schrodinger equation, we can get a differential equation in phi;
     e^-\frac{y^2}{2} \left[ \phi'' - 2y\phi' + (\epsilon - 1)\phi \right]= 0

    This is very similar to Hermite's Equation, and you can assume that there exists a power series solution, and get a recursion relationship from it that gives that, for n >= 0,
     a_{n+2} = a_{n} \frac{2n + 1 - \epsilon}{(n+1)(n+2)}
    From which the coefficients can be determined in the wavefunction.
    Now comes the part which you asked about; This series terminates when  \epsilon = 2n + 1 , and we know from what we set earlier that  E = \epsilon(\frac{1}{2}) \hbar \omega , and therefore  E = (n + \frac{1}{2}) \hbar \omega

    I'm not sure I've explained the last part very well, so ask me any questions if you want!
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by doomhalo)
    There's a few ways of approaching this, but it's quite a bit more Mathematically Heavy than for a rectangular problem.

    Step 1: De-dimensionalise
     \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}} + \frac{mx^2\omega^2}{2} \psi = E\psi

    We can change a few things in this problem to make it simpler; in this case we choose that  x = yl and then we substitute this into the equation; by de-dimensionalising it allows us to compare some quantities later which we wouldn't be able to do otherwise.

    Therefore, we get:

     \frac{-\hbar^{2}}{2ml^2} \frac{d^{2}\psi}{dy^{2}} + \frac{y^2 l^2 m \omega^2}{2} \psi = E\psi

    We multiply through by  \frac{-2ml^2}{\hbar^2}

    and then choose that  \frac{m^2 l^4 \omega^2}{\hbar^2} = 1
    which gives that  l = (\frac{\hbar}{m \omega})^{\frac{1}{2}}

    And choose that  \epsilon = \frac{-2 E m l^2}{\hbar^2}

    This simplifies our differential equation significantly; we get that

     \frac{d^2 \psi}{dy^2} - y^2 \psi = -\epsilon \psi

    Step 2: Look at the behaviour as y -> +/- infinity

    Outside of the potential, we need the wavefunction to die away, so we assume that  y^2 >> \epsilon and therefore we can say that
     \psi '' - y^2 \psi = 0 which has the solution
     \psi = e^{-\frac{1}{2}y^2}
    This accounts for the behaviour outside of the potential but not that inside; we say that  \Psi = \psi \phi_\epsilon and obviously we know psi. By substituting into the Schrodinger equation, we can get a differential equation in phi;
     e^-\frac{y^2}{2} \left[ \phi'' - 2y\phi' + (\epsilon - 1)\phi \right]= 0

    This is very similar to Hermite's Equation, and you can assume that there exists a power series solution, and get a recursion relationship from it that gives that, for n >= 0,
     a_{n+2} = a_{n} \frac{2n + 1 - \epsilon}{(n+1)(n+2)}
    From which the coefficients can be determined in the wavefunction.
    Now comes the part which you asked about; This series terminates when  \epsilon = 2n + 1 , and we know from what we set earlier that  E = \epsilon(\frac{1}{2}) \hbar \omega , and therefore  E = (n + \frac{1}{2}) \hbar \omega

    I'm not sure I've explained the last part very well, so ask me any questions if you want!
    Thank you very much for this. I took a long break from TSR and forgot about this thread.

    I was lacking a lot of the mathematical techniques back then but now I get it. Thanks for the reply.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 1, 2015
Poll
Do I go to The Streets tomorrow night?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.