# What are the energy states of a particle in a delta potential well $V(x)=-\delta(x)$?

I am trying to figure out how to derive the wavefunctions for a particle in a potential field $V(x)=-\delta(x)$.

Looking it up, I see that the actual energy states are of the form $e^{-\alpha |x|}$, but I don't understand how this is derived.

Here is my attempt. Schrodinger's equation asserts that: \begin{align} \hat{H}\psi &= E\psi \\ -\frac{\hbar^2}{2m}\psi'' - \delta &=E\psi. \end{align} Away from 0, this is just the free particle wave equation. Take $E_n$ to be the energy levels of this equation, and for brevity take $\alpha_n = \sqrt{\hbar^2/(2m)E_n}$. This means $\psi$ is some linear combination of $\psi_n$, each $\psi_n$ being of the form: \begin{equation} \psi_n=\left\lbrace \begin{matrix} A_1 e^{-i\alpha_n} + A_2 e^{i\alpha_n} & x < 0 \\ B_1e^{i\alpha_n} + B_2 e^{i\alpha_n} & x\geq 0\end{matrix}\right. \end{equation} with $A_i,B_i$ chosen so that $\psi_n$ is continuous and so that with some leniency, $\psi_n''=-\delta.$

Now say $\psi=\psi_n$ for some $n$. We would like to have $\langle \psi| \psi \rangle = 1,$ but no scale or choice of $A_1,A_2,B_1,B_2$ will make that integral even converge. Something is wrong...

Where is the error in this analysis and why don't I end up with energy states of the form $e^{|ax|}$?

## 2 Answers

You're very close. I think what you're missing is that you're looking for a bound state so $E<0$, therefore alpha is imaginary, making your exponentials in your superposition real exponential decays/growths instead of terms that oscillate off to infinity.

Two of these terms diverge towards $\pm\infty$, so the only way you're integral will converge is if the coefficients of those terms are zero. You're left with 2 coefficients, the "psi is continuous" requirement, and the normalization requirement. The integrals converge nicely, it's just math from here.

You're given that the potential is of form $V(x) = -\delta(x)$, and thus is centered at zero. Now, what you can do is to consider each region that the potential "separates" separately (namely region where $x>0$ and $x<0$), compute general solution with some coefficients, and find those coefficients using the definition of the potential. Consider region where ($x<0$), here the potential is exactly $0$, the Schodinger equation reduces to $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_1 = E\psi_1$$ The general solution of which is simply: $\psi_1 = Ae^{\kappa x}+Be^{-\kappa x}$ for $\kappa = \sqrt{-\frac{2mE}{\hbar^2}}$, but, since the potential is negative and for the energy eigenstate to be bound the energy must negative, the coefficient $\kappa$ is real: $\kappa\in\mathbb{R}$. For the wavefunction to be normalizable (and thus possible solution), $\lim_{x\rightarrow -\infty} \psi_1 = 0$ and thus $B=0$

Repeating the same process for the region where $x>0$, you obtain the following wavefunction: $\psi_2 = Ce^{\kappa x} + De^{-\kappa x}$. Similary, due to normalization, $C=0$ thus $\lim_{x\rightarrow \infty} \psi_2(x) = 0$

Now the question is, how to relate them?, well, you know the form of potential, so, the general form of Schrodinger equation for this problem is simply: $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) - \delta(x)\psi(x)= E\psi(x)$$ But since the potential is centered at zero (if fact has non-zero value only at $x=0$), you can integrate over the interval $[-\epsilon,\epsilon]$ as $\epsilon\rightarrow 0$: $$-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon} \frac{d\psi'}{dx} dx - \psi(0) \approx E\psi_{average}(x)\epsilon = 0$$ Which gives us discontinuity of the wavefunction's derivative: $$\psi'(\epsilon) - \psi'(-\epsilon) = -\frac{2m}{\hbar^2}\psi(0)\tag{D.1}$$ But you already know the general form of each solution in each region, specifically: $\psi_1(x)$ for $x<0$ and $\psi_2(x)$ for $x>0$, and the equation $\text{D.1}$ relates the derivatives of those functions. Computing the derivatives gives: $$\psi_1'(x) = A\kappa e^{\kappa x} \\ \psi_2'(x) = -D\kappa e^{-\kappa x}$$ Using discontinuity of derivative: $$\lim_{\epsilon\rightarrow 0} \psi'(\epsilon) - \psi'(-\epsilon) = \psi_2'(0) - \psi_1'(0) = -\frac{2m}{\hbar^2}\psi_1(0) = -\frac{2m}{\hbar^2}\psi_2(0)$$ But since $\psi_1(0) = A$, and $\psi_2(0) = B$, and from above equation (and continuity of wavefunction) $A=B$, let's only refer to $A$. Thus the above equation becomes: $$2\kappa A = \frac{2m}{\hbar^2}A \rightarrow \kappa = \frac{m}{\hbar^2}$$ Which gives you the energy eigenvalue for this bound wavefunction: $$E = -\frac{m}{2\hbar^2}$$ and the energy eigenstate is: $\psi_1(x) = Ae^{\kappa x}$ for $x<0$ and $Ae^{-\kappa x}$ for $x>0$ which is equivalent to the following: $$\psi(x) = Ae^{-\kappa |x|}$$ For since this equation has no zeros/nodes, this energy eigenstate is ground state. Since $n$-th energy eigenstate has $n-1$ zeros, $2$nd energy eigenstate must have exactly $1$ zero, lets consider what would be the consequence of $\exists x\in\mathbb{R}: \psi(x) = 0$. There are three possibilities, namely $\psi(x)=0$ at either

• $x=0$
• $-x\in\mathbb{R}^+$
• $x\in\mathbb{R}^{+}$

Lets consider each case separately: if $\psi(0)$ is zero, that would mean that the discontinuity of wavefunction's derivative (eqn. $\text{D.1}$) is equal to zero, which is equivalent to say: $\kappa=0 \rightarrow E=0$, i.e. particle doesn't exists, thus we can conclude that if zero exists, it is not located at $x=0$.

Consider the case when $-x\in\mathbb{R}^+$, or region where $x<0$, here the general solution for a wavefunction is $\psi_1(x) = Ae^{\kappa x} + Be^{-\kappa x}$. For it to have a zero in this region $Ae^{\kappa x}+ Be^{-\kappa x}=0$ for some $x$, which implies that $B\neq0$, and thus $$\lim_{x\rightarrow -\infty} \psi_1(x) = \lim_{x\rightarrow -\infty} Be^{-\kappa x} \neq 0$$ which prevents wavefunction from being normalizable.

Repeating similar process for region $x\in\mathbb{R}^+$ or $x>0$, gives the same answer: existence of zero implies non-normalizability of wavefunction, therefore for a bound wavefunction to exist, the following condition must hold $\forall x\in\mathbb{R}: \psi(x)\neq 0$, which proves that the ground state wavefunction is only possible bound energy eigenstate.

From above it is evident that for this potential there is only one bound energy eigenstate, namely $\psi(x)= Ae^{-|\kappa x|}$ with energy eigenvalue $E=-\frac{m}{2\hbar^2}$

Appendix: Solution for the proportionality coefficient $A$

Now, since we know the form of the groundstate wavefunction, we can compute the proportionality coefficient $A$ from normalization condition: $$\int_{-\infty}^{\infty} \overline{\psi}(x)\psi(x) dx = A^2\int_{-\infty}^{\infty} e^{-2|\kappa x|} dx =\\ A^2\left(\int_{-\infty}^{0} e^{2\kappa x} dx + \int_{0}^{\infty} e^{-2\kappa x} dx\right) = \frac{A^2}{\kappa} = 1$$ Which gives: $A = \kappa^{\frac{1}{2}} = \sqrt{\frac{m}{\hbar^2}}$, and the groundstate wavefunction takes the following form: $$\psi(x) = \sqrt{\frac{m}{\hbar^2}}e^{-|\kappa x|}$$

Note: to avoid overloading the text with derivatives, here any for any arbitrary function $f(x)$ of variable $x$, its derivative $\frac{df(x)}{dx}$ is denoted by $f'(x)$.