Delta function potential and Schrodinger Equation

In summary, the problem is that the second derivative of e^{-|x|} is not correct, and integrating both sides results in zero.
  • #1
Seriously
11
0
I have a time dependent wavefunction for inside a delta function potential well: V(x) = -a delta(x).

It reads
Psi(x,t) = (sqrt(m*a)/hbar) * exp(-m*a*abs(x)/hbar^2) * exp(-iEt/hbar)

I'm supposed to stick this back into the time dependent Schrodinger Equation and solve for E.

Taking my Psi(x,t), I found the second derivative with respect to x, and also found the time derivative. Then I plugged directly back into the time dependent Schrodinger Equation, with V(x) given as above. The problem is that I can't seem to make the delta function go away. I get
-(m*a^2)/(2*hbar^2) - a*delta(x) = E

How do I get the delta to go away? I can't just say x=0, because then the delta function is infinity.

I also tried: Plugging directly back into the time dependent Schrodinger Equation, I integrate both sides from -e to +e, where is some really small distance around x=0. Instead of using dPsi/dt, I write dPsi/dt in terms of d^2Psi/dx^2. Then I integrate both sides. The only problem with doing this, is when I integrate d^2Psi/dx^2 dx, this becomes dPsi/dx evaluated from -e to +e -- and because of the absolute value of x in Psi, this is zero. Help!
 
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  • #2
Where you went wrong was in taking the second derivative of [tex]e^{-|x|}[/tex]; it's actually a dirac delta, which cancels the other dirac deltas as you'd expect.

[tex]e^{-|x|}[/tex] is continuous and non-differentiable; the first derivative is has a jump discontinuity like the step function; and the derivative of that has a singularity like the dirac delta. (What you actually get is the sum of a dirac delta and a differentiable function.) A useful fun exercise for delta functions is problem 6 from Robert Jaffe.
 
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  • #3
Thank you for the reply, Rach3, I really appreciate it.

I see now that I'm not differentiating exp(-abs(x)) correctly, let alone the second derivative. But when I try to follow your instructions, I find that I'm still not comfortable with differentiating this.

Phi = exp(-abs(x))
dPhi/dx = - dabs(x)/dx exp(-abs(x))
where dabs(x)/dx = -1 for x<0, undef for x=0, +1 for x>0

Is the second derivative d2abs(x)/dx2 exp(abs(x))?
Or is it (-(dabs(x)/dx)^2 + d2abs(x)/dx2 ) exp(abs(x)) ?
Or none of the above, lol?
 

Related to Delta function potential and Schrodinger Equation

1. What is a Delta function potential?

A Delta function potential is a mathematical function used in quantum mechanics to describe a point-like interaction between a particle and a potential. It is often used to model interactions with an infinitely thin barrier or well.

2. How is the Delta function potential related to the Schrodinger Equation?

The Delta function potential is often used in the Schrodinger Equation to represent a localized interaction with a potential. It can be used to solve for the wavefunction and energy eigenvalues of a particle in the presence of a Delta function potential.

3. What are the properties of a Delta function potential?

A Delta function potential has the following properties:
- It is zero everywhere except at the point where it is defined.
- It is infinitely tall at the point where it is defined.
- The integral of the Delta function potential over all space is equal to one.

4. How does the strength of the Delta function potential affect the Schrodinger Equation?

The strength of the Delta function potential affects the Schrodinger Equation by changing the overall energy of the system. A stronger Delta function potential will result in a larger energy shift for the particle, while a weaker potential will result in a smaller energy shift. This can also affect the shape and behavior of the wavefunction.

5. Can the Delta function potential be used to describe real-life interactions?

No, the Delta function potential is a mathematical idealization and cannot be used to describe real-life interactions between particles. However, it is a useful tool in quantum mechanics for simplifying and solving certain problems and can be used as an approximation for more realistic potentials.

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