# An energy state

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.