Dirac Delta Potential

Classical Mechanics Level 5

Recall that the Dirac delta "function" (it is more properly a distribution) is thought of as a function $\delta(x)$ such that $\delta(x) = 0$ for all $x \neq 0$ , but such that $\int_{-\infty}^\infty f(x)\delta(x)\,dx \; = \; f(0)$ for all continuous functions $f$ . The Dirac delta potential below can be seen as an idealization of (and approximation for) a very deep, but very narrow square well potential $V_\varepsilon(x) \; = \; \left\{ \begin{array}{lll}-\tfrac{1}{2\varepsilon}V_0 & \hspace{1cm} & |x| < \varepsilon \\ 0 & & |x| > \varepsilon \end{array}\right.$ in the limit as $\varepsilon \to 0$ .

A one-dimensional quantum mechanical particle of mass $m$ moves in the presence of a potential $-V_0\delta$ , where $V_0 > 0$ and $\delta$ is the Dirac delta "function". Thus the wave function $\psi$ satisfies the time-independent Schrodinger equation $-\frac{\hbar^2}{2m} \frac{d^{2}\psi}{dx^2} - V_0\delta(x)\psi(x) \; = \; E\psi$ when the particle has energy $E$ .

There is only one bound state for this potential (there is only one value of $E$ which allows a nonzero solution $\psi$ of the wave equation such that $|\psi|^2$ over $\mathbb{R}$ exists). The bound state energy $E$ is of the form $E \; = \; \frac{amV_0^2}{\hbar^2}$ for some constant $a$ . What is the value of this constant?

The answer is -0.5.

1 solution

Mark Hennings
Feb 4, 2021

If $x \neq 0$ the wave equation states that

$\psi''(x) + \frac{2mE}{\hbar^2}\psi(x) \; = \; 0$

If $E > 0$ solutions of this equation are sinusoidal, and so will not give bound states. Thus $E < 0$ . If we write $E = -\frac{\hbar^2k^2}{2m}$ , then our bound state must be of the form $\psi(x) \; =\; \left\{ \begin{array}{lll}Ae^{-kx} & \hspace{1cm} & x >0 \\ Be^{kx} & & x < 0 \end{array}\right.$ for some constants $A,B$ . Certainly $\psi$ must be continuous, and so $A=B$ .

The singularity of the Dirac delta function means that $\psi'$ will not be continuous. Integrating the wave equation from $-u$ to $u$ gives

$\begin{aligned} -\frac{\hbar^2}{2m}\big[\psi'(u) - \psi'(-u)\big] - V_0\int_{-u}^u \delta(x)\psi(x)\,dx & = \; E\int_{-u}^u\psi(x)\,dx \\ \frac{\hbar^2Ak}{m}e^{-ku} - V_0A & = \; E\int_{-u}^u \psi(x)\,dx \; = \; \tfrac{2EA}{k}(1 - e^{-ku})\end{aligned}$

for any $u > 0$ . Letting $u \to 0+$ , we deduce that $\frac{\hbar^2Ak}{m} \; = \; V_0A$ and so, for a nonzero bound solution, we must have $k = \frac{mV_0}{\hbar^2}$ , and hence the bound state energy is $-\frac{\hbar^2k^2}{2m} \; = \; -\frac{mV_0^2}{2\hbar^2}$ which makes $a = \boxed{-\tfrac12}$ .

Very nice problem. I could not solve it. I tried approaching this using the Fourier transform as such:

$-P \psi''(x)-V_o \delta(x) \psi(x) = E\psi(x)$

The fourier transform definition used is:

$F_{\psi}(k) = \int_{-\infty}^{\infty}\psi(x) \mathrm{e}^{-ikx} \ dx$

where $P$ is the constant comprising of the reduced Planck's constant and mass.

Taking Fourier transform on both sides:

$-P(-k^2 F_{\psi}(k) ) - V_o \psi(0) = EF_{\psi}(k)$ $\implies F_{\psi}(k) =\frac{V_o \psi(0) }{Pk^2 - E}$

The wave function is therefore the inverse fourier transform of $F_{\psi}$ .

$\psi(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty}F_{\psi}(k) \mathrm{e}^{ikx} \ dk$

I just could not solve this beyond this point. I even tried a numerical route but the introduction of the Dirac delta makes the numerical problem messy.

Karan Chatrath - 4 months, 1 week ago

Your Fourier transform is not going to be square-integrable unless $E < 0$ , so the energy must be negative. Consider the Fourier transform of $e^{-a|x|}$ . That will get your answer, for the right value of $a>0$ , with the right negative energy...

Mark Hennings - 4 months, 1 week ago

I finally got the answer. I just did not consider the possibility of $E<0$ when reached the stage of inverse fourier transforming. Thank you for the tip.

I am thinking of how to solve your latest problem and it seems like I am out of my depth there as well. I am trying to find a solution to the time independent equation of the form:

$\psi(x) = \sum_{k=0}^{\infty} a_k x^k$

And then, I am trying to get relations for the coefficients $a_k$ . I don't know if this is correct as I seem to be getting nowhere at the moment.

Karan Chatrath - 4 months, 1 week ago

@Karan Chatrath – If you are going to do it that way, let me give you a few hints. Start looking at the Hamiltonian without the $\tfrac{1}{\sqrt{2}}x$ term (just $\tfrac12(P^2+Q^2)$ ). Differential equation theory would then suggest trying solutions of the form $e^{-\frac12x^2}f(x)$ . Try writing $f(x)$ as a power series expansion, and obtain a recurrence relation for the coefficients. For some values of $E$ , the power series for $f(x)$ terminates, and $f(x)$ becomes a polynomial. That leaves you with two questions: (1) why is termination of the power series necessary to give bound state solutions, (2) how do you incorporate the extra term into the Hamiltonian?

Mark Hennings - 4 months, 1 week ago

@Mark Hennings – I have started slowly working out the steps as you have suggested. Thank you for the suggestions. I will try and post a solution if I succeed, despite you having done so already.

Karan Chatrath - 4 months ago

Nice solution, Mr Mark Hennings and nice question. Do you remember me ? Hot Integral guy haha! You always post solution on my questions! Hope you are fine

Aman Rajput - 1 month, 2 weeks ago

Dirac Delta Potential

The answer is -0.5.

1 solution

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