Time independent Schrodinger equation

The Schrodinger equation in a single dimension can be reduced to two ordinary differential equations through separation of variables. This yields one equation that is dependent on time, and one that is time independent. It is also known as the Energy Eigenvalue Equation.

$$-\frac{{\hslash }^2}{2m}\frac{\partial \Psi (x,t)}{\partial x^2}+V(x)\Psi (x,t)=i\hslash \frac{\partial \Psi (x,t)}{\partial t}$$

Above is the Schrodinger Equation. It can be solved for a free particle. See (i.e no potential energy, $V(x)$ term is zero).

When $V(x)$ is independent of time, we can separate the Schrodinger equation into

$$\frac{df(t)}{dt}=-\frac{iE}{\hslash }f(t)$$

and

$$-\frac{{\hslash }^2}{2m}\frac{\partial \psi (x,t)}{\partial x^2}+V(x)\psi (x)=E\psi (x)$$

The latter is known as the Time independent Schrodinger equation, while the former can be rather easily solved using an ansatz of $F(t)=f(0)e^{-iEt/\hslash }$ , or $F(t)=f(0)e^{i\omega }$ where $E=\hslash \omega$ .

NB:

$$|\Psi (x,t){|}^2=|\Psi (x){|}^2$$

Probability is Independent of time

Using the Time independent Schrodinger equation, and a suitable $V(x)$ we can model a “particle in a box”, also known as an Infinite square well. We can also use the Time independent Schrodinger equation to model Step Potentials as seen later on.

Three Dimensional Versions

The Three dimensional Time independent Schrodinger equation is given by

$$-\frac{{\hslash }^2}{2m}\left[\frac{d^2\psi }{d{x}^2}+\frac{d^2\psi }{d{y}^2}+\frac{d^2\psi }{d{z}^2}\right]=(E-V(r))\psi (x,y,z)$$

in Cartesian coordinates. NB- This can also be represented with the Laplacian operator ${\nabla }^2$.

The Laplacian in spherical coordinates is given below: $-\frac{{\hslash }^2}{2m}\left[\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)+\frac{1}{r^2\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d}{d\theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{d^2}{d{\varphi }^2}\right]\psi (r,\theta ,\varphi )=(E-V(r))\psi (r,\theta ,\varphi )$ This gives rise to the momentum operator in three dimensions: $\mathbf{P}$, as well as the more interesting angular momentum operator ${\mathbf{L}}_{op}$, which will be covered more in next class.

$${\mathbf{P}}^2=-{\hslash }^2{\nabla }^2$$ $$\mathbf{L}=-{\hslash }^2\left[\frac{1}{\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d}{d\theta }\right)+\frac{1}{\sin \theta }\frac{d^2}{d{\varphi }^2}\right]=\mathbf{r}\times \mathbf{p}$$

The rotational kinetic energy term in the Hamiltonian (RHS of following) is also equal to the angular momentum over $2m{r}^2$

$$\frac{{\mathbf{L}}_{op}^2}{2m{r}^2}=-\frac{{\hslash }^2}{2m}\left[\frac{1}{r^2\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d}{d\theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{d^2}{d{\varphi }^2}\right]$$

Separation of Variables in Spherical Coordinates

Replacing the Hamiltonian operator with $\left(\frac{{\hat{P}}_r^2}{2m}+\frac{{\hat{L}}^2}{2m{r}^2}\right)$, we get

$$\left(\frac{{\hat{P}}_r^2}{2m}+\frac{{\hat{L}}^2}{2m{r}^2}\right)\Psi (r,\theta ,\varphi )=(E-V(r))\Psi (r,\theta ,\varphi )$$

Replacing $\Psi$ with $R(r)Y(\theta ,\varphi )$

$$⟹\left(\frac{{\hat{P}}_r^2}{2m}+\frac{{\hat{L}}^2}{2m{r}^2}\right)R(r)Y(\theta ,\varphi )=(E-V)R(r)Y(\theta ,\varphi )$$