# 1D Potential Step QM Problem

1. Oct 15, 2013

### CAF123

1. The problem statement, all variables and given/known data
Consider a one-dimensional quantum system described by the potential: $$V(x) = -V_o + \frac{1}{2}mw^2x^2\,\,,V_o > 0\,\,\text{for}\,\, |x| < b\,\,\text{and}\,\,0\,\,\text{otherwise}$$
Show that the state described by: $$\psi_{-}(x) = R_{-} \exp\left[-\frac{x^2}{2x_o^2}\right]\,\,\,\text{for}\,\,|x| < b$$
can be an eigenstate of the Hamiltonian describing a bound state (i.e. a state with
negative energy) only for a specific value of $x_o$.

2. Relevant equations
$\hat{H}\psi_{-}(x) = E\psi_{-}(x)$ if $\psi_{-}$ is an eigenstate of the Hamiltonian.

3. The attempt at a solution
I have sketched the potential and it looks parabolic in the region $x \in (-b,b)$ A bound state is a state of neg energy => E < 0 and so the path of some incoming particle intersects the potential plot in two places. For $x \in (-b,b)$ it makes sense that the wave function is a decaying exponential. Because $V_o$ is fixed, it also makes sense that the decaying exp is not proportional to m or w, since I think this means there is only one parabola that satisfies the boundary conditions. If $\psi_{-}$ is to be an eigenstate, it must satisfy the eigenvalue problem $\hat{H}\psi_{-} = E \psi_{-}$. I tried subbing $\psi_{-}$ in, and this gave me an expression for E in terms of x. I am not really sure what else to try, any hints would be great.
Many thanks.

Last edited: Oct 15, 2013
2. Oct 15, 2013

### Staff: Mentor

??
Then you should try other values of x_0, too.

Okay
Which path, which incoming particle?

Why?
(it does not, and the wave function is not a decaying exponential)

Where is the relation between the statements here?

Good.
How does this expression look like?
How does it have to look like for an eigenstate?

3. Oct 15, 2013

### CAF123

I am not sure what you mean - if x is not in (-b,b) then V is zero. If x is in (-b,b) then the potential is parabolic with a minimum value Vo

For the purposes of the question, the particle has -Vo< E < 0. $\psi_{-}$ must be associated to a quantum particle which has negative energy. I am only trying to get an idea of what is going on, I don't know if it helps me solve the question or not.
For the particle to be in (-b,b) the particle must have penetrated one side of the potential barrier. Is $\psi_{-}$ not a decaying exponential?

Is this the right approach? Subbing in $\psi_{-}$into the TISE: $-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi_{-} + V(x)\psi_{-} = E\psi_{-}$ gives the following; $$\psi_{-} \left[\frac{\hbar^2}{2m} \frac{1}{x_o^2} - \frac{\hbar^2}{2m} \frac{x^2}{x_o^6} + \frac{1}{2}mw^2x^2 - V_0\right] = E\psi_{-}.$$ For $\psi_{-}$ to be an eigenstate, the bracketed part must equal E. But why does this only give one $x_o$?

Last edited: Oct 15, 2013
4. Oct 15, 2013

### Staff: Mentor

x_0 means x0, that is not x.
Did you plot the function for different values of x0? Try $x_0=\frac{b}{5}$, for example. Do you see a deviation from a parabola?

The energy scale is arbitrary, there is no special meaning of "negative energy" - apart from the fact that this is the energy range where you are supposed to look for solutions.

It is not. It would be a decaying exponential outside of (-b,b) if (!) the potential there would be constant - it is not.

It is.

The important point here: it must be E independent of x. How can the bracket be independent of x?

5. Oct 16, 2013

### CAF123

I think I see your point - at x =-b,b, you could have a spike. However, a parabolic form is still okay I think.

Okay I see now that it is obviously not a decaying exponential, a Gaussian rather. Outside the region [-b,b], the potential is a constant (=0), so should I not expect a decaying solution? When I solve the Schrodinger eqn this is what I get?

So that means the coefficient of $x^2$ in the bracketed part must be zero. I.e $$\frac{1}{2} mw^2 - \frac{\hbar^2}{2m} \frac{1}{x_0^4}= 0 \Rightarrow x_0^4 = \frac{\hbar^2}{m^2 w^2}$$ and so $x_o = \pm \frac{\hbar^{1/2}}{w^{1/2}m^{1/2}}$.

The next part of the question is about finding the wave function for all x and the probabilities of finding the particle in or outside the well. I think it makes more sense to consider the particle already inside the well - I think if you consider instead an incoming particle then in this set up, it would have negative kinetic energy which is not physical. For x < -b and x > b, the Schrodinger eqn is $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi_{-} = E \psi_{-}\Rightarrow \frac{d^2}{dx^2} \psi_{-} = -\frac{2mE}{\hbar^2}\psi_{-}$$For a bound state, E < 0 and so $-2mE/\hbar^2$ will be positive. Hence $\psi_{-} = A\exp(\sqrt{2mE}/\hbar x)$ for x < -b and $\psi_{-} = D\exp(-\sqrt{2mE}/\hbar x)$ for x > b. (Reject the diverging exp terms) By the continuity of the wave function I can determine A and D in terms of $R_{-}$. Does this seem ok?

Last edited: Oct 16, 2013
6. Oct 16, 2013

### Staff: Mentor

I would not call this a spike, but it can be discontinuous. And I misinterpreted your potential, see below:

Inside, okay.
Oh sorry, I misread the potential. Okay, outside you get a decaying exponential, I agree.

Good. Both x0 give the same wave function.

There is no incoming particle.

The way you use the signs looks wrong. If E<0, your square root is imaginary, and you don't get a decay.

7. Oct 17, 2013

### CAF123

I am not really sure how I would correct this. The negative that appears there in the first place (in the h/2m term) is already there in the Schrodinger eqn. So I will definitely have a negative on the RHS. Then I said that since E is negative in our case, then this makes the RHS positive. So when I find the roots of the characteristic polynomial I have real solutions.

Last edited: Oct 17, 2013
8. Oct 17, 2013

### Staff: Mentor

The minus sign should stay in the square root, as (-2mE) is positive and will give real roots, while (2mE) is/does not. Just check the individual steps, it has to be there.