Question

Schrodinger time independent wave equation derivation

Answer 1

$\Huge \textsf{Schr\"odinger's Equation}$

Schrödinger's Famous Equation, in one dimension, is:

$\displaystyle i\hbar \frac{\partial \Psi(x,t)}{\partial t}=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\hat{V}\right)\Psi(x,t) \\ \\ \\ \implies \boxed{i\hbar \frac{\partial \Psi(x,t)}{\partial t} = \hat{H}\Psi(x,t)}$

Here, $\Psi(x,t)$ is the Wave-Function and $\hat{H}$ is the Hamiltonian Operator. We want to solve the equation and derive the Time-Independent Equation.

We see that on the Left Hand Side, we have a partial derivative with respect to time, while on the right hand side, we have a double partial derivative with respect to the position x.

We cannot solve this Partial Differential Equation directly, since $\Psi(x,t)$ is a function of both position and time.

So, our strategy is to assume $\boldsymbol{\Psi}\bold{(x,t)}$ to be a product of two independent functions, one function of position and the other function of time.

So we write:

$\boldsymbol{\Psi(x,t)=\psi(x).f(t)}$

And so we can now proceed to solving:

$\displaystyle i\hbar \frac{\partial \Psi(x,t)}{\partial t} = \hat{H}\Psi(x,t)\\\\\\\implies i\hbar \frac{\partial \psi(x).f(t)}{\partial t} = \hat{H}\psi(x).f(t)$

On the LHS, $\psi(x)$ acts as a constant and comes out, since derivative is only with respect to time. Similarly, on the RHS, $f(t)$ comes out as a constant.

$\displaystyle \implies i\hbar \, \psi(x)\frac{\partial f(t)}{\partial t} = f(t) \, \hat{H}\psi(x) \\\\\\\implies \frac{i\hbar}{f(t)}\frac{\partial f(t)}{\partial t}=\frac{1}{\psi(x)}\hat{H}\psi(x)$

Now, we see that The LHS is completely dependent on time, and the RHS is completely dependent on Position. This is possible if both LHS and RHS are equal to some constant.

Suppose that the constant is k.

Then,

$\displaystyle \frac{i\hbar}{f(t)}\frac{\partial f(t)}{\partial t}=\frac{1}{\psi(x)}\hat{H}\psi(x)=k$

Consider the following part.

$\displaystyle \frac{1}{\psi(x)}\hat{H}\psi(x)=k \\\\\\\implies \hat{H}\psi(x)=k\,\psi(x)$

We see that when the Hamiltonian operator operates on $\psi(x)$, we get the same function $\psi(x)$ back.

So, $\boldsymbol{\psi(x)}$ is an eigenfunction of the Hamiltonian Operator.

Now, there is a Postulate of Quantum Mechanics, which states that:

For a Linear and Hermitian Operator acting on the wave function, the eigenvalue is an observable quantity related to the operator.

Here, $\hat{H}$ is the Hamiltonian Operator, which is essentially the sum of Kinetic Energy and Potential Energy Operators.

So, the observable related to the Hamiltonian Operator is Energy (E).

So, we write the eigenvalue as k = E.

Hence, our equation becomes:

$\displaystyle \hat{H}\psi(x)=k\, \psi(x) \\ \\ \\ \implies \huge \boxed{\bold{\hat{H}\,\boldsymbol{\psi}(x)=E\,\boldsymbol{\psi}(x)}}$

This is the Time-Independent Schrödinger's Equation.