Solving the Schrodinger Equation for V(x)=A sech^2(αx)

[Mahasweta]

1. How can I solve the Schrodinger equation for a potential V(x)= A sech^2(αx) ? How do I come to know that whether sech(αx) is a non-node bound state of the particular or not?




2. p^2/2m + V(x) = E



3. exp(kx)[A tanh(αx) + C]

 

[Simon Bridge]

Welcome to PF;

How can I solve the Schrodinger equation for a potential V(x)= A sech^2(αx) ?

You put the potential into the Schodinger equation with appropriate boundary conditions - just like any DE.
Note: $$\text{sech}(x)=\frac{2e^{-x}}{1+e^{-2x}}$$

How do I come to know that whether sech(αx) is a non-node bound state of the particular or not?

... "non bound state of a particular" what? That sentence is incomplete.

i.e. are you saying that you are given $\psi=\text{sech}(ax)$ and you want to know if it is the wavefuction of a bound energy eigenstate of the potential you've been given, if it is a bound state of any potential or what?

You can figure out a lot about a potential by plotting it and using your experience of solving for different wells - like what sorts of potentials have bound states etc.

 

[Mahasweta]

I meant that for a particular potential how do I come to know that among a set of wave functions for that potential which one is non-node bound state?

 

[Simon Bridge]

Well you have two conditions to be satisfied here.
1. the state is bound
2. the state has no nodes

Do you know how to test for these conditions separately?
Do you know what these conditions mean?

Perhaps this will help?
http://arxiv.org/pdf/quant-ph/0702260.pdf

 

[paranormal]

The Schrodinger equation for a potential V(x)= A sech^2(αx) can be solved using various techniques, such as the shooting method, the WKB approximation, or numerical methods. However, the most common and straightforward approach is to use the method of separation of variables.

To solve the Schrodinger equation, we first need to express it in the form of a differential equation. Using the given potential and the general form of the Schrodinger equation, we get:

-ħ^2/2m d^2ψ/dx^2 + A sech^2(αx) ψ = Eψ

Next, we can use the substitution ψ(x) = u(x) exp(ikx) to transform the equation into a simpler form, where u(x) is a function of x and k is a constant:

-ħ^2/2m d^2u/dx^2 + (k^2 - α^2)u = 0

This is a second-order ordinary differential equation, which can be solved using standard techniques. The solution will depend on the value of k, which is related to the energy of the system. By solving for u(x) and substituting it back into the original equation, we can obtain the wavefunction ψ(x) and the corresponding energy eigenvalue E.

To determine whether sech(αx) is a non-node bound state or not, we can look at the behavior of the wavefunction at large values of x. If the wavefunction decays to zero as x approaches infinity, then it is a non-node bound state. This can be seen from the general solution of the Schrodinger equation, which has the form:

ψ(x) = exp(kx)[A tanh(αx) + C]

For a non-node bound state, the term in the brackets must approach zero as x goes to infinity, which means that A must be equal to zero. Therefore, if A is non-zero, then sech(αx) is not a non-node bound state.

In summary, to solve the Schrodinger equation for a potential V(x)= A sech^2(αx), we can use the method of separation of variables. The resulting differential equation can then be solved to obtain the wavefunction and energy eigenvalue. To determine if sech(αx) is a non-node bound state, we