# Form of Solution to Schrodinger Equation

• mrrelativistic
In summary, the form of the wave function for the Schrodinger equation of a finite potential well and potential barrier is determined by solving the time-independent Schrodinger equation in each region and making the wave function and its derivative continuous at the boundaries. For a potential barrier, the wave function takes the form of Ae^ikx + Be^-ikx (x < -a), Fe^ikx (x > a), and Ce^μx + De^-μx (-a < x < a). For a potential well, the wave function is Ae^kx + Be^-kx (x < -a), Fe^kx + Ge^-kx (x > a), and Csin(μx)
mrrelativistic
Hey guys, I just want to ask on how do you determine the form of wave function for Schrodinger equation of finite potential well and potential barrier.

Why is it ψ(x) = Ae^ikx + Be^-ikx (x < -a)
ψ(x) = Fe^ikx (x > a)
ψ(x) = Ce^μx + De^-μx (-a < x < a)
for k^2 = (2mE)/(hbarr), μ^2 = (2m(V - E))/(hbarr) and E < V

The potential barrier is 0 at x < -a and x > +a and V at -a < x < a.

I know that there are two cases (E < V and E > V). What I really don't know is how do you get the solution form of the Schrodinger equation.

And for a potential well:

Why is it ψ(x) = Ae^kx + Be^-kx (x < -a)
ψ(x) = Fe^kx + Ge^-kx (x > a)
ψ(x) = Csin(μx) + Dcos(μx) (-a < x < a)

The potential well is 0 at x < -a and x > +a and -V at -a < x < a.

mrrelativistic said:
Hey guys, I just want to ask on how do you determine the form of wave function for Schrodinger equation of finite potential well and potential barrier.
You [STRIKE]guess[/STRIKE] select a suitable ansatz (see http://en.wikipedia.org/wiki/Ansatz), then insert your [strike]guess[/STRIKE] carefully chosen ansatz back into the equation to see if it's a good solution.

That's how many differential equations are solved.

Well, in this case there's no need to guess the solution but you can derive it directly from the time-independent Schrödinger equation, which obviously is the one the OP likes to solve. Setting $\hbar=1$ for convenience, it reads
$$-\frac{1}{2m} \psi''(x)+V(x) \psi(x)=E \psi(x),$$
where $E \in \mathbb{R}$ is an eigenvalue of the Hamilton operator.

To find the possible eigenvalues, you have to solve this equation and find the values for $E$ such that the resulting wave functions become either square-integrable (leading to the discrete part of the Hamiltonian's spectrum and the corresponding the bound-state solutions of the time-independent Schrödinger equation) or functions that are normalizable to $\delta$ distributions (continuous part of the Hamiltonian's spectrum and scattering solutions of the time-independent Schrödinger equation).

Here, the potential is piece-wise constant and it's pretty easy to solve the differential equation in the three regions separately. Then you have to make the wave function and its first derivative continuous at the places where the potential makes (a) jump(s).

These conditions give unique (generalized) eigenvalues $E$ and the associated (generalized) eigenfunctions!

at start you want to separate 3 different regions because you got different V potential as you can see in the picture.So generaly you solve 3 Schrodinger equations for the 3 regions.
The results must ofcourse unite to 1 wave function keeping in mind that a wave function(and its derivative)must be continuous.

Anyway The first thought is how much energy you start with, as you said there are the tow cases of E < V and E > V

for E<V

you got in Schrodinger for the 3 regions

1st region x<-L ) (-h^2/2m)ψ1''+V*ψ1=Eψ1 ψ1''=d^2ψ(x)/dx^2 , h=h(barr)

2cond -L<x<L ) (-h^2/2m)ψ2''=Eψ2

3rd x>L ) (-h^2/2m)ψ3''+V*ψ3=Eψ3

So now we solve them

1st) ψ1''=(2m/h^2)(V-E)ψ1 , (2m/h^2)>0 ,(V-E)>0 using k^2=(2m/h^2)(V-E) you get ψ1''=k^2ψ1 solve→ ψ1(x)= Ae^kx + Be^-kx

for that solution if we take x → -∞ we get ψ1=∞ and you don't expect that cause you are in the potential region and its pushing you away, or you can say that you want ψ to be Square-integrable function , so B=0

2cond) (-h^2/2m)ψ2''=(2m/h^2)(-E)ψ2 now (2m/h^2)(-E)<0 using γ^2=(2m/h^2)(E) you get
ψ2''=-γ^2 ψ2 solve→ ψ2(x) = Ce^iγx + De^-iγx

now to get the form you said use Eulers Eq and you ll get ψ(x) = Vsin(γx) + Mcos(γx)
(mind V and M are now complex)

3rd) solve same way as the first with same k^2
and you get ψ(x) = Fe^kx + Ge^-kx (x > a) for the same reason you don't expct to get too big
So F=0

after that to get the whole solution you just use boundary conditions.

Now for E>V you get different kind of problem witch is scatering one.The difference applies essentially that now we don't worry if ψ is Square-integrable function

mrrelativistic said:
how do you determine the form of wave function for Schrodinger equation of finite potential well and potential barrier.

Have you had a course in differential equations yet? The time-independent SE in each region is a linear homogeneous differential equation which can be solved using the method of "characteristic polynomials." A Google search on "linear differential equation characteristic polynomial" will give you several sites that describe this method.

In practice, physicists usually use Nugatory's "guess and try it" method first. If that fails, we use a more formal method like characteristic polynomials, or the power-series method (which you might see when you get to the SE for the simple harmonic oscillator), etc.

Oh, I see. I could solve it or I could guess and test it if it satisfies the Schrodinger equation for that region. Thanks Nugatory. vanhees71, SyLba, and jtbell. Really helpful answers.

## 1. What is the Schrodinger Equation?

The Schrodinger Equation is a fundamental equation in quantum mechanics that describes the behavior of quantum systems over time. It was developed by Austrian physicist Erwin Schrodinger in 1926 and is used to calculate the wave function, which describes the quantum state of a system.

## 2. What is the significance of solving the Schrodinger Equation?

Solving the Schrodinger Equation allows us to predict the behavior of quantum systems, such as the location and energy of electrons in an atom. It also helps us understand the underlying principles of quantum mechanics and has led to many important discoveries and applications in fields such as chemistry, materials science, and electronics.

## 3. What is the form of the solution to the Schrodinger Equation?

The form of the solution to the Schrodinger Equation depends on the specific system being studied. In general, the solution is a wave function, which is a mathematical function that describes the probability of finding a particle in a certain location or state. The form of the wave function can vary depending on factors such as the potential energy of the system and the dimensions of the space it occupies.

## 4. How is the Schrodinger Equation solved?

The Schrodinger Equation can be solved using various mathematical techniques, such as separation of variables, perturbation theory, and numerical methods. The specific method used depends on the complexity of the system being studied and the desired level of accuracy in the solution.

## 5. What are some real-world applications of the Schrodinger Equation?

The Schrodinger Equation has numerous applications in modern technology. It is used in the design of semiconductors for electronics, the development of new materials for energy storage and conversion, and the creation of quantum computers. It also plays a crucial role in understanding and predicting chemical reactions and the behavior of particles in accelerators and nuclear reactors.

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