# Finding Stationary Wavefunction with a Line Potential

• doggydan42
In summary, the particle has a potential:$$V(x) = \begin{cases} V_0 & x > 0 \\ 0 & x \leq 0 \end{cases}$$and can be in one of four states. There are no initial conditions, only boundary conditions.
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##.

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 ?

BvU said:
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.

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

##\psi'## should be continuous.

BvU said:
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.

doggydan42 said:
this still gives a superposition of the two functions
Almost correct... : a combination of two functions in different domains
doggydan42 said:
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 ?

Last edited:
BvU said:
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:
$$Ak = A'\kappa-B'\kappa$$
$$B = -\frac{k}{\kappa}$$
$$A' + B' = -\frac{k}{\kappa}$$

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

BvU said:
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:
$$A = 1, B = -\frac{k}{\kappa}, A' = 0, B' = \frac{-k}{\kappa}$$

doggydan42 said:
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)

Last edited:

## 1. What is the concept of a stationary wavefunction?

A stationary wavefunction is a mathematical representation of the state of a quantum system. It describes the probability of finding a particle in a specific location or energy state. A stationary wavefunction does not change over time, hence the term "stationary".

## 2. What is a line potential in quantum mechanics?

A line potential is a type of potential energy function used in quantum mechanics to describe the behavior of a particle in a one-dimensional system. It represents the energy of a particle as a function of its position along a single line or axis.

## 3. How do you find the stationary wavefunction with a line potential?

To find the stationary wavefunction with a line potential, you need to solve the Schrödinger equation for the given potential. This involves finding the eigenvalues and eigenfunctions of the Hamiltonian operator, which represents the total energy of the system. The stationary wavefunction is the eigenfunction associated with the lowest energy eigenvalue.

## 4. What are some common line potentials used in quantum mechanics?

Some common line potentials used in quantum mechanics include the infinite square well potential, the harmonic oscillator potential, and the step potential. These potentials are often used to model physical systems such as atoms, molecules, and particles confined to a specific region.

## 5. Why is finding the stationary wavefunction important in quantum mechanics?

Finding the stationary wavefunction is important because it allows us to make predictions about the behavior of quantum systems. By knowing the probability of finding a particle in a particular location or energy state, we can determine the properties and behavior of a system. This is crucial in understanding the fundamental principles of quantum mechanics and its applications in various fields of science and technology.

Replies
5
Views
1K
Replies
2
Views
1K
Replies
10
Views
613
Replies
29
Views
468
Replies
7
Views
2K
Replies
10
Views
707
Replies
16
Views
852
Replies
16
Views
2K