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Quantum Mechanics: Delta function potential next to an infinitely tall potential barrier

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data

    We learned in class that a particle exposed to a 1D delta-function potential well would
    always have a single bound state. Let us now explore this question for the case where the
    delta-function potential well is situated in the vicinity of the impenetrable potential wall:
    Given the potential energy:

    [itex]V(x)=\begin{cases}
    +\infty,\quad\text{for}\quad x<0\\
    -C\delta(x-a),\quad\text{for}\quad 0<x<\infty
    \end{cases}[/itex]

    Explore the formation of a bound state for a particle of mass m by solving the time independent
    Schrödinger equation. Here [itex]a > 0[/itex] specifies the position of the potential well
    and [itex]C > 0[/itex] its strength.
    (a) Find the (transcendental) equation that would allow you to determine the energy of the
    bound state. Indicate how you would be solving this equation graphically.
    (b) In the absence of the neighboring wall (think if this may correspond to the limit [itex]a\rightarrow\infty [/itex]),
    the bound state is formed for any value of strength [itex]C[/itex]. But now, do we have a restriction
    on the value of [itex]C[/itex] for a bound state to exist? If so, what is the minimal possible value
    of [itex]C[/itex] in terms of other given parameters of the problem?
    (c) Can you bring simple physical arguments to rationalize the results in item (b)?

    2. Relevant equations

    Eqn 1):[itex]\frac{d^2\psi(x)}{dx^2}-\frac{2m}{\hbar^2}[V(x)-E]\psi(x)=0[/itex]

    3. The attempt at a solution

    Part a):

    First since the well is centered at [itex]x=a[/itex] I'll split the [itex]x[/itex]-axis into two regions:

    Region 1: [itex]0<x<a[/itex]

    Region 2: [itex]x>a[/itex]

    Next since I'll setup my finite well of strength [itex]C[/itex]:

    Let the well have a width of [itex]2\varepsilon[/itex]. Then it's depth is [itex]\frac{C}{2\varepsilon}[/itex]. So [itex] C [/itex] must have dimensions of (J)(m). Let [itex] C=V_0(2\varepsilon) [/itex] Where the well has a maximum depth of [itex]-V_0[/itex]. The well has its walls at [itex]x=a\pm\varepsilon[/itex]. In order for a bound state to exist [itex]E<0[/itex]. So there are two cases to consider within the well:

    Case I: A particle with energy [itex]E[/itex]

    [itex]\frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}[E]\psi(x)=0[/itex]

    Let [itex]\kappa^2=-\frac{2mE}{\hbar^2}[/itex]

    [itex]\frac{d^2\psi(x)}{dx^2}-\kappa^2\psi(x)=0[/itex]

    Which has solutions of [itex]C_1e^{\kappa x}[/itex]and[itex]C_2e^{-\kappa x}[/itex]

    Case II: A particle with energy [itex] -V_0-E[/itex]

    [itex]\frac{d^2\psi(x)}{dx^2}-\frac{2m}{\hbar^2}[-V_0-E]\psi(x)=0[/itex]

    Let [itex] k^2=\frac{2m}{\hbar^2}[V_0+E][/itex]

    [itex]\frac{d^2\psi(x)}{dx^2}+k^2\psi(x)=0[/itex]

    which has solutions of [itex]C_3\cos(kx)[/itex]and[itex]C_4\sin(kx)[/itex]

    Now I can impose my boundary conditions:

    In Region 1; [itex]\psi_{I}(0)=0[/itex]
    [itex]\psi_{I}(a-\varepsilon)=\psi_{II}(a-\varepsilon)[/itex]
    [itex]\psi'_{I}(a-\varepsilon)=\psi'_{II}(a-\varepsilon)[/itex]

    In Region 2; [itex] \lim_{x\rightarrow\infty}\psi_{I}(x)=0[/itex]
    [itex]\psi_{I}(a+\varepsilon)=\psi_{II}(a+\varepsilon)[/itex]
    [itex]\psi'_{I}(a+\varepsilon)=\psi'_{II}(a+\varepsilon)[/itex]

    Let [itex]a'=a-\varepsilon[/itex] and [itex] a''=a+\varepsilon[/itex]

    Now to solve the TISE in Region 1:

    [itex]\psi_{I}(x)=C_1e^{\kappa x}+C_2e^{-\kappa x}[/itex]

    [itex]\psi_{II}(x)=C_3\cos(kx)+C_4\sin(kx)[/itex]

    [itex]\psi_{I}(0)=0=C_1+C_2\Rightarrow C_2=-C_1[/itex]

    [itex]\psi_{I}(x) = C_1[e^{\kappa x}-e^{-\kappa x}][/itex]

    Now I'll select an even function from [itex]\psi_{II}(x)[/itex](Letting [itex]C_4=0[/itex]). Such that [itex]\psi_{II}(x)=C_3\cos(kx)[/itex]

    Using my B.Cs:

    [itex]\psi_{I}(a')=\psi_{II}(a')[/itex]

    Eqn 2:[itex] C_1[e^{\kappa a'}-e^{-\kappa a'}]=C_3\cos(ka')[/itex]

    [itex]\psi'_{I}(a')=\psi'_{II}(a')[/itex]

    Eqn 3:[itex] C_1\kappa[e^{\kappa a'}+e^{-\kappa a'}] = -C_3k\sin(ka') [/itex]

    Dividing Eqn 3 by Eqn 2:

    Eqn 4: [itex]\kappa\frac{e^{\kappa a'}+e^{-\kappa a'}}{e^{\kappa a'}-e^{-\kappa a'}}=-k\tan(ka')[/itex]

    or equivalently

    Eqn 4:[itex]\kappa a'\tanh(\kappa a')=-ka'\tan(ka')[/itex]

    Now I'll select an odd function from [itex]\psi_{II}(x)[/itex](Letting [itex]C_3=0[/itex]). Such that [itex]\psi_{II}(x)=C_4\sin(kx)[/itex]

    [itex]\psi_{I}(a')=\psi_{II}(a')[/itex]

    Eqn 5:[itex] C_1[e^{\kappa a'}-e^{-\kappa a'}]=C_4\sin(ka')[/itex]

    [itex]\psi'_{I}(a')=\psi'_{II}(a')[/itex]

    Eqn 6:[itex] C_1\kappa[e^{\kappa a'}+e^{-\kappa a'}] = C_4k\cos(ka') [/itex]

    Dividing Eqn 3 by Eqn 2:

    Eqn 7: [itex]\kappa\frac{e^{\kappa a'}+e^{-\kappa a'}}{e^{\kappa a'}-e^{-\kappa a'}}=k\cot(ka')[/itex]

    or equivalently

    Eqn 7:[itex]\kappa a'\tanh(\kappa a')=ka'\cot(ka')[/itex]

    Now to solve the TISE in Region 2:

    [itex]\psi_{I}(x)=C_5e^{\kappa x}+C_6e^{-\kappa x}[/itex]

    [itex]\psi_{II}(x)=C_7\cos(kx)+C_8\sin(kx)[/itex]

    [itex]\lim_{x\rightarrow\infty}\psi_{I}(x)=0\Rightarrow C_5=0[/itex]

    [itex]\psi_{I}(x)=C_6e^{-\kappa x}[/itex]

    Again I'll start with the even solution [itex]\psi_{II}(x)=C_7\cos(kx)[/itex]

    [itex]\psi_{I}(a'')=\psi_{II}(a'')[/itex]

    Eqn 8:[itex]C_6e^{-\kappa a''}=C_7\cos(ka'')[/itex]

    [itex]\psi'_{I}(a'')=\psi'_{II}(a'')[/itex]

    Eqn 9: [itex]-C_6\kappa e^{-\kappa a''}=-C_7k\sin(ka'')[/itex]

    Dividing Eqn 9 by Eqn 8:

    Eqn 10: [itex] \kappa a''=ka''\tan(ka'')[/itex]

    Now for the odd portion of [itex] \psi_{II}(x)[/itex].

    [itex]\psi_{I}(a'')=\psi_{II}(a'')[/itex]

    Eqn 11:[itex]C_6e^{-\kappa a''}=C_8\sin(ka'')[/itex]

    [itex]\psi'_{I}(a'')=\psi'_{II}(a'')[/itex]

    Eqn 12: [itex]-C_6\kappa e^{-\kappa a''}=C_8k\cos(ka'')[/itex]

    Dividing Eqn 12 by Eqn 11:

    Eqn 13: [itex] -\kappa a''=ka''\cot(ka'')[/itex]

    I also have that:

    Eqn 14: [itex] \kappa^2+k^2=\frac{2mV_0}{\hbar^2}[/itex]

    [itex]\kappa=\sqrt{\frac{2mV_0}{\hbar^2}-k^2}[/itex]

    Eqn 15: [itex]\kappa a'=\sqrt{\frac{2mV_0(a')^2}{\hbar^2}-(ka')^2}[/itex]

    Eqn 16: [itex]\kappa a''=\sqrt{\frac{2mV_0(a'')^2}{\hbar^2}-(ka'')^2}[/itex]

    At this point I'm a little lost as to what I need to do to get a single transcendental equation; however, here is how I think I should proceed:

    1) for all equations sub in the expressions for [itex]a'[/itex] and [itex]a''[/itex] in terms of [itex] a [/itex]
    2) use eqns 15 and 16 to get an expression for [itex]\kappa a[/itex]
    3) for regions 1 and 2 plot odd and even solutions in terms of [itex] \kappa a[/itex]
    (Edit: actually I just realized since [itex]\varepsilon[/itex] is small I can take [itex]\varepsilon^2[/itex] to be [itex]0[/itex] so what I'll end up with is just [itex]\kappa a = \sqrt{\frac{V_0}{\frac{\hbar^2}{2ma^2}}-(ka)^2}[/itex]) I'm still not sure how to resolve the transcendental equations in regions 1 and 2 though. Thanks in advance!
     
    Last edited: Sep 28, 2014
  2. jcsd
  3. Sep 28, 2014 #2

    Orodruin

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    Instead of treating the delta distribution as a limit of a square potential, I suggest using it as it is and infer from this the boundary condition that must be fulfilled (in addition to continuity) by the wave function at x = a. You will then only have one matching to perform.
     
  4. Sep 28, 2014 #3
    I'm not sure how to deal directly with the delta function potential. The way we dealt with such a potential in class was to construct such a finite square well. Though I'm curious to hear more about your approach. Thank you for the reply :)
     
  5. Sep 28, 2014 #4

    vela

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    The idea is to integrate the Schrodinger equation from ##a-\varepsilon## to ##a+\varepsilon## and take the limit as ##\varepsilon \to 0^+##. That'll give you the condition ##\psi'(x)## has to satisfy at the boundary.
     
  6. Sep 28, 2014 #5
    Thank you I'll reformulate my problem and post my solution later :)
     
  7. Sep 29, 2014 #6
    [itex] \lim_{\varepsilon\rightarrow0}\int_{x=a-\varepsilon}^{a+\varepsilon}\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}-C\delta(x-a)\psi(x)dx=\lim_{\varepsilon\rightarrow0}\int_{x=a-\varepsilon}^{a+\varepsilon}E\psi(x)dx [/itex]

    [itex] \lim_{\varepsilon\rightarrow0}[\psi'(a+\varepsilon)-\psi'(a-\varepsilon)]=\frac{2mC}{\hbar^2}\psi(a)[/itex]

    Now in region 1: [itex]\psi_{I}(x)=C_1e^{\kappa x}+C_2e^{-\kappa x}[/itex]

    applying the B.C in region 1 of [itex]\psi_{I}(0)=0\Rightarrow C_2=-C_1[/itex]

    Then [itex]\psi_{I}(x)=C_1(e^{\kappa x}-e^{-\kappa x})[/itex]

    Now for region 2: [itex] \psi_{I}(x)=C_3e^{-\kappa x} [/itex]

    which satisfies in region 2: [itex] \lim_{x\rightarrow\infty}\psi_{I}(x)=0[/itex]

    Now [itex]\psi_{I}(x)[/itex] must be continuous at [itex]x=a[/itex] so

    [itex] C_1(e^{\kappa a}-e^{-\kappa a})=C_3e^{-\kappa a} [/itex]

    [itex] C_1=C_3\frac{e^{-\kappa a} }{e^{\kappa a}-e^{-\kappa a}} [/itex]
     
    Last edited: Sep 29, 2014
  8. Sep 29, 2014 #7
    In region 1: [itex]\psi'_{I}(a-\varepsilon)=C_3\frac{e^{-\kappa a}}{e^{\kappa a}-e^{-\kappa a}}\kappa(e^{\kappa(a-\varepsilon)}+e^{-\kappa(a-\varepsilon)})[/itex]

    In region 2: [itex] \psi'_{I}(a+\varepsilon)=-C_3\kappa e^{-\kappa a}[/itex]

    So [itex]\lim_{\varepsilon\rightarrow0}[\psi'_{I}(a+\varepsilon)-\psi'_{I}(a-\varepsilon)]=-\frac{2mC}{\hbar^2}C_3e^{-\kappa a}[/itex]

    Which becomes [itex] \lim_{\varepsilon\rightarrow0}[(-C_3\kappa e^{-\kappa (a+\epsilon)})-(C_3\frac{e^{-\kappa a}}{e^{\kappa a}-e^{-\kappa a}}\kappa(e^{\kappa(a-\varepsilon)}+e^{-\kappa(a-\varepsilon)}))]=-\frac{2mC}{\hbar^2}C_3e^{-\kappa a}[/itex]

    [itex][(-C_3\kappa e^{-\kappa a})-(C_3\frac{e^{-\kappa a}}{e^{\kappa a}-e^{-\kappa a}}\kappa(e^{\kappa a}+e^{-\kappa a}))]=-\frac{2mC}{\hbar^2}C_3e^{-\kappa a}[/itex]

    Finally,

    [itex][(\kappa)+(\frac{1}{e^{\kappa a}-e^{-\kappa a}}\kappa(e^{\kappa a}+e^{-\kappa a}))]=\frac{2mC}{\hbar^2}[/itex]

    Which as [itex] a\rightarrow\infty [/itex] gives me [itex] \kappa = \frac{mC}{\hbar^2} [/itex]
     
  9. Sep 30, 2014 #8

    vela

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    Looks good. You could simplify the algebra a bit if you used
    \begin{eqnarray*}
    \psi_I(x) &= A\sinh ka \\
    \psi_{II}(x) &= Be^{-ka}
    \end{eqnarray*} In terms of hyperbolic trig functions, your transcendental equation is
    $$k(1+\coth ka) = \frac{2mC}{\hbar^2}.$$
     
  10. Sep 30, 2014 #9
    Awesome thanks for help everyone! :) Cheers
     
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