Delta function potential
From Wikipedia, the free encyclopedia
In quantum mechanics the delta function potential provides a simple example of a situation that can be solved with the Schrödinger equation. It is a one-dimensional potential and is given by:
That is, it is a potential well which is zero everywhere except at x = 0.
Contents |
[edit] Bound solution
When the potential is of the form above, the solution of the Schrödinger equation shows that the bound state wavefunction is:
And the energy can have only one value, and that is:
[edit] Derivation of bound solution
We are interested in finding the wave function for a bound state of this delta function potential. The wave function will be a solution to the Schrödinger equation:
-
where
is the mass of the particle is the (complex valued) wavefunction that we want to find is a function describing the potential at each point x and 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 Schrödinger equation will be a different function on each side:
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 Schrödinger equation takes the form:
Rearanging this,
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:
The solution to differential equations of this form involve the exponential:
-
- where in this situation,
We can eliminate the second term in ψ1 because in this left-hand region it would blow up as x approaches .
So, in this region,
[edit] Right side of the potential
In the region to the right of the potential, the Schrödinger equation takes on the same form in the previous section,
The only difference now is we have to keep the exponentially decreasing term - the other one would blow up as x approaches .
So,
[edit] Energy of the bound state
So far, it has been shown that the solution must have the form
-
- .
- where
- , 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, , must be continuous except where the potential blows up (goes to infinity).
Putting the first rule into math:
Therefore,
The last trick, is to integrate the Schrödinger equation around x=0, but do so over a really small range:
-
- where ε is a really small number.
The right-hand side is
and this is a more accurate approximation the smaller ε is. In the limit of this just goes to 0.
The left-hand side is
Which, after rearanging is
-
- .
Let ε get closer and closer to zero, and remember that to the left of the delta potential ψ = ψ1 and on the right of it ψ = ψ2:
Thus we have a relationship between the constant c in the wavefunction and the "strength" of the delta function potential:
But don't forget that c is related to the energy, so plugging that in,
This condition shows the one allowed energy, which is:
[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):
This has to equal one, so
This can be plugged back into the wavefunction and we find:
-
- .
Which can be written more compactly as
[edit] See also
- The free particle
- The particle in a box
- The finite potential well
- The particle in a ring
- The particle in a spherically symmetric potential
- The quantum harmonic oscillator
- The hydrogen atom or hydrogen-like atom
- The ring wave guide
- The particle in a one-dimensional lattice (periodic potential)
[edit] References
- Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0-13-805326-X.