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I am trying to do this quantum mechanics exercise:

A particle with mass $m$ in the potential:

$$V(x)= \left\{\begin{array}{lc}\infty& x\leq -a\\ \Omega\, \delta(x) & x>-a\end{array}\right.$$

I have to find the energy eigenfunction and eigenvalue.

I imposed the boundary conditions:

$$\psi(0^-)-\psi(0^+)=0$$ $$\psi'(0^-)-\psi'(0^+)=0$$

where a prime stands for the $x$-derivative. But according to the solution the correct condition on the $\psi'$ is:

$$\psi'(0^-)-\psi'(0^+)=\dfrac{2m\,\Omega}{\hbar^2}\,\psi(0)$$

Why is it so?

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    $\begingroup$ This is a very weird potential. Are you sure of it? $\endgroup$ – QuantumBrick Apr 3 '17 at 19:32
  • $\begingroup$ @QuantumBrick Yes I am sure. $\endgroup$ – mattiav27 Apr 3 '17 at 20:01
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    $\begingroup$ Have a look at Griffith's Introduction to Quantum Mechanics section 2.5: the delta function potential.. Equation 2.107 (in my book) covers your boundary condition. $\endgroup$ – user129412 Apr 3 '17 at 20:13
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    $\begingroup$ I don't see how this question is homework like. The asker wants to know why the weird boundary condition in presence of the delta function, which seems like a perfectly reasonable question to me. $\endgroup$ – Javier Apr 4 '17 at 16:09
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Let's first look at an "intuitive" explanation. Take a look a the Schrödinger equation:

$$-\frac{\hbar^2}{2m} \psi'' + V(x)\psi = E\psi$$

The wavefunction can't have any delta functions in it, because it must be square integrable. Therefore, there can't be any delta functions in the right hand side, and the $V\psi$ term has a delta function in it, so $\psi''$ must have a delta function to cancel the one from the potential: if $\psi''$ has a delta function, then $\psi'$ has an infinitely high derivative at $x=0$, which means that it is discontinuous.

How to calculate the jump? Integrate both sides of the equation from $-\epsilon$ to $\epsilon$; we will let $\epsilon$ go to zero at the end. The first term is

$$ -\frac{\hbar^2}{2m} \int_{-\epsilon}^\epsilon dx\ \psi'' = -\frac{\hbar^2}{2m} (\psi'(\epsilon) - \psi'(-\epsilon)) \to -\frac{\hbar^2}{2m} \Delta \psi',$$

where $\Delta \psi'$ is the jump in $\psi'$, which is what we're looking for.

The potential term is

$$\int_{-\epsilon}^\epsilon dx\ \Omega \delta(x) \psi(x) = \Omega \psi(0)$$

And the last term just goes to zero as $\epsilon \to 0$. From this you can get the boundary condition.

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