*Part 6 in the quest for the hydrogen molecule*

In this part I will present, as promised, a derivation of the famous Schrödinger equation.

As I have shown in the last part, with our approach so far we are able to calculate important properties of physical systems already. Usually these properties are calculated from the Schrödinger equation, which is assumed at the start to do quantum mechanics. To show you that this is not a necessary starting point, I will now do the opposite: I will derive the Schrödinger equation from our assumption that \(\langle x|p\rangle=e^{ixp/\hbar}\).

Consider a particle in any kind of position-dependent potential \(V(x)\). The energy of the particle is

$$E=\frac{p^2}{2m} + V(x).$$

We write down the energy equation to determine the energy states \(|\psi\rangle\):

$$\hat E |\psi\rangle = E |\psi\rangle.$$

This is all just the same as we did in the previous part. This time we will project the energy equation on position states \(\langle x|\):

$$\langle x|\hat E |\psi\rangle = E\langle x|\psi\rangle=E\psi(x),$$

where we define a function \(\psi(x) \equiv \langle x|\psi\rangle\). Again, we split the equation into the kinetic and potential energy terms:

$$\langle x|\hat E |\psi\rangle = \langle x|\hat E_k |\psi\rangle+\langle x|\hat E_p|\psi\rangle.$$

## Kinetic energy

The kinetic energy term of the energy equation needs a little bit of mathematics. The trick is to insert a projection on momentum states \(|p\rangle\) *and* a projection on position states \(|x’\rangle\):

$$\langle x|\hat E_k |\psi\rangle=\langle x| \frac{\hat{p}^2}{2m} |\psi\rangle=\\

\int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ \langle x| \frac{\hat{p}^2}{2m}| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle.$$

We can now remove the hats from the operators \(\hat p\) and start filling in the known expressions for \(\langle x|p\rangle\) and \(\langle x|\psi\rangle\):

$$\langle x| \frac{\hat{p}^2}{2m}| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle=\\

\frac{p^2}{2m} \langle x| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle=\\

\frac{p^2}{2m} e^{ip(x-x’)/\hbar}\psi(x’).$$

Now comes the main trick: I integrate over \(x’\) using integration by parts twice. First time:

$$\frac{p^2}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi(x’)=\\

\frac{i\hbar p}{2m} \bigg[e^{ip(x-x’)/\hbar}\psi(x’)\bigg]_{-\infty}^{+\infty}-

\frac{i\hbar p}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi'(x’),$$

where \(\psi'(x)\) denotes the derivative of the function \(\psi(x)\). We reason that the value of \(\psi(x)\) at \(x = \pm \infty\) should be 0 since it should not be physically relevant for the problem. We can always put the system we are studying in a very large box from which the particles cannot escape, and it should give the same results. Therefore I drop the first term. I continue with the second integration by parts, again dropping the \(\psi'(x)|_{-\infty}^{+\infty}\) term:

$$-\frac{i\hbar p}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi'(x’)=\\

-\frac{\hbar^2}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi^″ (x’).$$

To get rid of the two integrals, I can take the reverse steps of what I did before, replacing the integrals by the projection on states and dropping that projection since it doesn’t really do anything.

$$-\frac{\hbar^2}{2m} \int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi^″ (x’)=\\

-\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ \langle x|p\rangle \langle p| x’\rangle\psi^″ (x’)=\\

-\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty}\! dx’ \langle x|x’\rangle\psi^″ (x’)=\\

-\frac{\hbar^2}{2m} \psi^″ (x),$$

where the last step is explained in part 5.

## Potential energy

This time, the potential energy term is the easiest:

$$\langle x|\hat E_p|\psi\rangle=\langle x|V(\hat{x})|\psi\rangle.$$

As we established in part 5, an operator like \(\hat{x}\) may also operate on the projected states on the left, so we can continue with

$$\langle x|V(\hat{x})|\psi\rangle=

V(x)\langle x|\psi\rangle=

V(x)\psi(x).$$

## Results

Putting the kinetic and potential energy terms together again, we find

$$-\frac{\hbar^2}{2m} \psi^″ (x)+V(x)\psi(x)=E \psi(x),$$

which is called the time-independent Schrödinger equation.