Delta function potential

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In quantum mechanics the delta function potential provides a simple example of a situation that can be solved with the Schrodinger equation. It is a one-dimensional potential and is given by:

$V(x) = -a \delta (x) \,$

That is, it is a potential well which is zero everywhere except at $x = 0$ .

1 Bound solution
2 Derivation of bound solution
3 See also
4 References

[edit] Bound solution

The graph of the bound state wavefunction solution to the delta function potential is continuous everywhere, but it's derivative is not at x=0.

When the potential is of the form above, the solution of the Schrodinger equation shows that the bound state wavefunction is:

$\psi = \frac{\sqrt{a m}}{\hbar} e^{-ma|x|/\hbar^2}$

And the energy can have only one value, and that is:

$E = \frac{-m a^2}{2 \hbar^2}$

[edit] Derivation of bound solution

We are interesting in finding the wave function for a bound state of this delta function potential. The wave function will be a solution to the Schrodinger equation:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} + V(x) \psi = E \psi$	where $\hbar = \frac{h}{2 \pi}$
	$m \,$ is the mass of the particle
	$\psi\,$ is the (complex valued) wavefunction that we want to find
	$V\left(x\right)\,$ is a function describing the potential at each point x and
	$E\,$ is the energy, a real number.

In this case V is this delta function potential mentioned above, and it divides the space into two regions. As it will turn out, the solution to the Schrodinger equation will be a different function on each side:

$\psi (x) = \begin{cases} \psi_1 (x) & \mbox{for} \ \ x<0, \\ \psi_2 (x) & \mbox{for} \ \ 0<x \end{cases}$

But one should expect a symmetrical answer for a bound state, and that will be shown to be true.

[edit] Left side of the potential

In the region to the left of the delta function, the potential is zero so the Schrodinger equation takes the form:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi_1}{d x^2} = E \psi_1$

Rearanging this,

$\frac{d^2 \psi_1}{d x^2} + \frac{2 m E}{\hbar^2} \psi_1 = 0$

Now, for a bound state the energy, E, of the particle must be less than zero. So to make it more clear what the solution will be, write this as:

$\frac{d^2 \psi_1}{d x^2} - \frac{2 m |E|}{\hbar^2} \psi_1 = 0$

The solution to differential equations of this form involve the exponential:

$\psi_1 = A e^{cx} + B e^{-cx} \,$

where in this situation,

$c = \frac{\sqrt{2m|E|}}{\hbar}$

We can eliminate the second term in $ψ 1$ because in this left-hand region it would blow up as x approaches $-\infty$ .

So, in this region,

$\psi_1 = A e^{cx} \,$

[edit] Right side of the potential

In the region to the right of the potential, the Schrodinger equation takes on the same form in the previous section,

$\frac{d^2 \psi_2}{d x^2} - \frac{2 m |E|}{\hbar^2} \psi_2 = 0$

The only difference now is we have to keep the exponentially decreasing term - the other one would blow up as x approaches $\infty$ .

So,

$\psi_2 = B e^{-cx} \,$

[edit] Energy of the bound state

So far, it has been shown that the solution must have the form

$\psi (x) = \begin{cases} A e^{cx} & \mbox{for} \ \ x<0, \\ B e^{-cx} & \mbox{for} \ \ 0<x \end{cases}$ .

where

$c = \frac{\sqrt{2m|E|}}{\hbar}$ , and

A and B are yet to be determined.

One can find out more about the solution by applying two important rules about wavefunctions:

A wavefunction, $ψ$ , must be continuous everywhere and
Its derivatives, $\frac{d\psi}{dx}$ , must be continuous except where the potential blows up (goes to infinity).

Putting the first rule into math:

$\psi_1 (x=0) = \psi_2 (x=0) \,$

$A e^{0} = B e^{-0} \,$

Therefore,

$A = B \,$

The last trick, is to integrate the Schrodinger equation around x=0, but do so over a really small range:

$-\frac{\hbar^2}{2 m} \int_{-\epsilon}^{\epsilon} \frac{d^2 \psi}{d x^2}\, dx + \int_{-\epsilon}^{\epsilon}V(x) \psi \, dx = E \int_{-\epsilon}^{\epsilon} \psi \, dx$

where

ε

is a really small number.

The right-hand side is

$E \int_{-\epsilon}^{\epsilon} \psi \, dx \approx E \cdot 2 \epsilon \cdot \psi (x=0)$

and this is a more accurate approximation the smaller $ε$ is. In the limit of $\epsilon \to 0$ this just goes to 0.

The left-hand side is

$-\frac{\hbar^2}{2 m} \left( \frac{d\psi}{dx}\bigg|_{\epsilon} - \frac{d\psi}{dx}\bigg|_{-\epsilon} \right) -a \int_{-\epsilon}^{\epsilon}\delta(x) \psi \, dx = 0$

Which, after rearanging is

$\frac{d\psi}{dx}\bigg|_{\epsilon} - \frac{d\psi}{dx}\bigg|_{-\epsilon} = -\frac{2 m a}{\hbar^2} \psi (0)$ .

Let $ε$ get closer and closer to zero, and remember that to the left of the delta potential $ψ = ψ 1$ and on the right of it $ψ = ψ 2$ :

$\psi_2' (0) - \psi_1' (0) = -\frac{2 m a}{\hbar^2} \psi (0)$

$-A c - A c = -\frac{2 m a}{\hbar^2} A$

Thus we have a relationship between the constant c in the wavefunction and the "strength" of the delta function potential:

$c = \frac{m a}{\hbar^2}$

But don't forget that c is related to the energy, so plugging that in,

$\frac{m a}{\hbar^2} = \frac{\sqrt{2m|E|}}{\hbar}$

This condition shows the one allowed energy, which is:

$E = \frac{-ma^2}{2 \hbar^2}$

[edit] Normalization

Finally, to figure out what the constant A out front of the wavefunction is, one can normalize the wavefunction (i.e. make sure the probability of finding the particle somewhere equals 1):

$\int_{-\infty}^{\infty} \psi^* \psi \, dx = 2 A^2 \int_{0}^{\infty} e^{-2 c x} \, dx = \frac{2 A^2}{2 c}$

This has to equal one, so

$A = \sqrt{c} = \frac{\sqrt{ma}}{\hbar}$

This can be plugged back into the wavefunction and we find:

$\psi (x) = \begin{cases} \frac{\sqrt{ma}}{\hbar} e^{\frac{max}{\hbar^2}} & \mbox{for} \ \ x<0, \\ \frac{\sqrt{ma}}{\hbar} e^{-\frac{max}{\hbar^2}} & \mbox{for} \ \ 0<x \end{cases}$ .