Finding Stationary Wavefunction with a Line Potential

[doggydan42]

Homework Statement

A particle of mass m in one dimension has a potential:
$$V(x) =
\begin{cases}
V_0 & x > 0 \\
0 & x \leq 0
\end{cases}
$$

Find $\psi(x)$ for energies $0 < E < V_0$, with parameters
$$k^2 = \frac{2mE}{\hbar^2}$$
and
$$\kappa^2 = \frac{2m(V_0 - E)}{\hbar^2}$$

Use coefficients such that $\psi(0) = -\frac{k}{\kappa}$

Homework Equations

The time-independent schrodinger equation:
$$\hat H \psi(x) = E \psi(x)$$

The Attempt at a Solution

I started with the potential being 0, which gave.
$$\psi''(x) = -k^2\psi(x)$$
This would give solutions in the form
$$\psi(x \leq 0) = Asin(kx) + Bcos(kx)$$
For the potential $V_0$, I got
$$\psi''(x) = -\kappa^2\psi(x)$$
Similarly, $$\psi(x > 0) = Asin(\kappa x) + Bcos(\kappa x)$$

I was not sure how to choose coefficients.
I was thinking that for $x \leq 0$, I could find $B = -\frac{k}{\kappa}$. Though I could not figure out what to do for A, and for $x > 0$.

Thank you in advance

 

[BvU]

With $0<E<V_0$, one of your $k^2$ is < 0, the other > 0. How do you solve the Schroedinger equation for each half of hte axis, exactly ?

 

[doggydan42]

With $0<E<V_0$, one of your $k^2$ is < 0, the other > 0. How do you solve the Schroedinger equation for each half of hte axis, exactly ?

Both $k^2$ and $\kappa^2$ should be greater than 0. $k^2 = 2mE/hbar^2$ and $\kappa^2 = 2m(V_0-E)/hbar^2$ E is greater than 0, and $V_0 - E > 0$. To solve the wavefunction, I used the potential for each half. For one half, I just substituted 0. For the other, I substituted $V_0$. These gave me a solution in the form of a superposition of waves. I have one boundary condition to find one of the coefficients, the coefficient of the cosine should be $-k/\kappa$ for both x > 0 and x <= 0. My issue was finding the coefficient for the sine. There should be some boundary condition using $\psi'(x)$, but this was not given, and I did not know how to go about finding it.

 

[BvU]

Show how you solve the Schoedinger equation for $x>0$. In detail, in particular wrt the signs.

$\psi'$ should be continuous.

 

[doggydan42]

Show how you solve the Schoedinger equation for $x>0$. In detail, in particular wrt the signs.

$\psi'$ should be continuous.

I used $\hat H \psi = E \psi$. Using $V(x) = V_0$:
$$-\frac{\hbar^2}{2m}\psi''(x) + V_0 \psi(x) = E\psi(x)$$
$$\psi''(x) -\frac{2m}{\hbar^2}V_0\psi(x) = -\frac{2m}{\hbar^2} E\psi(x)$$
$$\psi''(x) = \frac{2m}{\hbar^2}V_0\psi(x)-\frac{2m}{\hbar^2} E\psi(x) = \frac{2m}{\hbar^2} (V_0 - E)\psi(x)$$
Assuming a solution in the form $\psi(x) = e^{\lambda x}$ gives $\lambda = \pm \kappa$.
I see where I went wrong with my initial solution; however, this still gives a superposition of the two functions:
$$\psi(x > 0) = Ae^{\kappa x}+ Be^{-\kappa x}$$.
The initial condition would give that $A+B = -\frac{k}{\kappa}$. Another condition is needed to solve for A and B.

 

[BvU]

this still gives a superposition of the two functions

Almost correct... : a combination of two functions in different domains

The initial condition

There are no initial conditions, only boundary conditions: $\psi$ and $\psi'$ have to be continuous at $x=0$.

How many integration constants do you actually have ? Any further conditions ?

 

[doggydan42]

Almost correct... : a combination of two functions in different domains
There are no initial conditions, only boundary conditions: $\psi$ and $\psi'$ have to be continuous at $x=0$.

How many integration constants do you actually have ? Any further conditions ?

The conditions given are that $\psi(0) = -\frac{k}{\kappa}$ and that it is not normalizable. Since V(x) has a discontinuity but not a delta-function, $\psi'$ and $\psi$ are continuous, as you have stated.

Using the first condition:
For $x \leq 0$,
$$\psi(0) = Asin(0) + Bcos(0) = B = -\frac{k}{\kappa}$$
For $x > 0$,
$$\psi(0) = A'e^0 + B' e^0 = A' + B' = -\frac{k}{\kappa}$$, where A' and B' are the coefficients for $x > 0$ to be distinct from $x \leq 0$.
Since $\psi'$ is continuous, $\psi'(0)$ is the same for $x \leq 0$ and $x > 0$.
So,
For $x \leq 0$,
$$\psi'(0) = Akcos(0)-Bksin(0) = Ak$$
For $x > 0$,
$$\psi'(0) = A'\kappa-B'\kappa$$

I need one more equation, since there are 4 variables. I have 3 equations for these four variables:
$$ Ak = A'\kappa-B'\kappa$$
$$B = -\frac{k}{\kappa}$$
$$A' + B' = -\frac{k}{\kappa}$$

 

[BvU]

Good you put quotes behind two of the constants. I can deduce you have four integration constants. Two conditions pus a forced value for $\psi(0)$ leaves one opening. What about the behaviour of $\psi$ for $x\rightarrow \pm\infty$ ?

We come to the conclusion there are no normalizable steady state solutions (hence the $\psi(0)$ condition).

neveretheless the case is of substantial interest; cf a free particle: $\psi(x)$ can not be normalized there either.
https://en.wikipedia.org/wiki/Solution_of_Schr%C3%B6dinger_equation_for_a_step_potential

 

[doggydan42]

Good you put quotes behind two of the constants. I can deduce you have four integration constants. Two conditions pus a forced value for $\psi(0)$ leaves one opening. What about the behaviour of $\psi$ for $x\rightarrow \pm\infty$ ?

We come to the conclusion there are no normalizable steady state solutions (hence the $\psi(0)$ condition).

neveretheless the case is of substantial interest; cf a free particle: $\psi(x)$ can not be normalized there either.
https://en.wikipedia.org/wiki/Solution_of_Schr%C3%B6dinger_equation_for_a_step_potential

Should the behavior of $\psi$ for $x \rightarrow \infty$ be 0?

If so, then 0 = B', since for $x \rightarrow \infty$, $e^{\kappa x} \rightarrow \infty$ and $e^{-\kappa x} \rightarrow 0$.
Using A' = 0, $B' = \frac{-k}{\kappa}$, and $Ak = -\frac{-k}{\kappa}\kappa = k$, so $A = 1$.
Overall:
$$A = 1, B = -\frac{k}{\kappa}, A' = 0, B' = \frac{-k}{\kappa}$$

 

[BvU]

Should the behavior of $\psi$ for $x \rightarrow \infty$ be 0?

yes, it should go to zero fast enough to make it quadratically integrable.
[edit: but see #8]

Others use a C instead of a B' (look at his case II, p 20 ff)