*Part 4 in the quest for the hydrogen molecule*

In this part I would like to focus on the energy of a particle. A particle can have kinetic and potential energy. Suppose a particle is in a state \(|\psi\rangle\) where it has energy \(E\), then how does that state look like? Is there only one such state or are there many? What are the position or momentum properties of such a state?

We have already seen that the position and the momentum of a particle are one and the same property. Interestingly, the energy of a particle will typically depend on its position and momentum. So the goal is to find a state that describes the position and the momentum properties of the particle so that it has energy \(E\).

It turns out to be very useful to be able to write down in a formula the statement that “the value of property Q of state \(|\psi\rangle\) is equal to q”. It is written like this:

$$\hat Q |\psi\rangle = q |\psi\rangle$$

We have put a little hat above the \( \hat Q \) to denote that it measures the value of the property \(Q\) of the state that is to the right of it, and we call \( \hat Q \) an **operator**. Note that since this is quantum mechanics, not all properties of a state are necessarily well-defined. For example, the momentum property \(\hat p \)of a state \(|p\rangle\) is well known:

$$\hat p |p\rangle = p |p\rangle,$$

but the position property \(\hat x \) of that state:

$$\hat x |p\rangle = ???$$

is not really clear (at the moment).

Let’s get started with some actual physical results! To start things simple, we will take the simplest case of a free particle in one dimension. For a free particle, its energy is given by:

$$E=\frac{1}{2}mv^2 = \frac{p^2}{2m}.$$

As required, the energy of \(|\psi\rangle\) must be \(E\), we will put this down in an equation we will call the **energy equation**:

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

Now we just apply the procedure we have seen before: we know that \(|\psi\rangle\) can be written as a superposition of position states\(|x\rangle\) or of momentum states \(|p\rangle\). Since the energy of the particle is related to its momentum, we choose the latter:

$$| \psi \rangle = \int_{-\infty}^{+\infty} \psi(p) |p \rangle dp,$$

So the question now becomes, what is the value of \(\psi(p)\)? We break down both the left and right side in the energy equation as a superposition of momentum states:

$$\hat E \int_{-\infty}^{+\infty} \psi(p) |p \rangle dp = E \int_{-\infty}^{+\infty} \psi(p) |p \rangle dp.$$

As stated before, we don’t know what \(\hat E\) does with state \(|p \rangle\), but no worries, we will replace it with the free-particle energy operator \(\frac{\hat p^2}{2m}\) that measures its energy in terms of momentum!

$$\hat E \int_{-\infty}^{+\infty} \psi(p) |p \rangle dp =\\

\int_{-\infty}^{+\infty} \psi(p) \hat E |p \rangle dp =\\

\int_{-\infty}^{+\infty} \psi(p) \frac{\hat p^2}{2m} |p \rangle dp =\\

\int_{-\infty}^{+\infty} \psi(p) \frac{ p^2}{2m} |p \rangle dp.$$

Taking the last line and putting it back into the left side of the energy equation,

$$\int_{-\infty}^{+\infty} \frac{ p^2}{2m} \psi(p) |p \rangle dp=

\int_{-\infty}^{+\infty} E \psi(p) |p \rangle dp,$$

we can conclude that, in order for the left and right sides to be equal, the integrand has to be equal for every value of \(p\):

$$ \frac{ p^2}{2m}\psi(p) = E \psi(p).$$

We find that \(\psi(p)\) has to be \(0\) whenever \(p^2 \neq 2mE\). On the other hand, for \(p_-=-\sqrt{2mE}\) and \(p_+=\sqrt{2mE}\), \(\psi(p)\) can take any value we like. Let’s call those values \(A\) and \(B\), so that we conclude that the states

$$|\psi\rangle=A|p_-\rangle+B|p_+\rangle$$

are the states with energy \(E\). This may not seem too impressive to you, since what we find is basically what we expected: a state of a particle with momentum \(p_-=-\sqrt{2mE}\), or a state of a particle with momentum \(p_+=\sqrt{2mE}\), or any combination thereof, are the states with energy \(E\). But we have done a proper quantum-mechanical derivation of this result, and that procedure will come in handy in the next installment.