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.

Properties and observables

Part 3 in the quest for the hydrogen molecule

Last time we talked about how the position and velocity of a particle are not independent properties. In fact, they are totally dependent! Unfortunately this is sometimes explained as that they can not be independently ‘measured’ or that there is ‘uncertainty’. That is absolute rubbish! Quantum mechanics is the most exact and predictive theory there is. However, one cannot extract information from reality that does not exist in that reality. We have to let go of our classical tendencies, and venture into the quantum world and become enlightened.

There are different ways we can postulate this and they are all unfortunately a little magical. Usually textbooks will talk about ‘conjugate variables’ or the ‘uncertaintly principle’, but I would like to try a different approach. It is equally unprovable, but I think it is a rather practical approach, and it will be fun to try to deduce the entire theory of quantum mechanics from it. However we will also have to start doing mathematics now, so let’s get to work!

Let us first focus on superpositions again. Take the case of a particle with a position in one dimension, and take a superposition like this:

$$| \psi \rangle = 0.2 | x=3 \rangle + 0.9| x=5 \rangle.$$

We have said before that a state is the full description of the situation and it can be a superposition of pure states. Then any state describing the position property of a particle is a superposition of the pure states of that particle being at specific locations \(x\).

$$| \psi \rangle = 0.2 | x=0 \rangle + 0.01| x=1 \rangle+0.5| x=2 \rangle+0.4| x=3 \rangle +\ldots.$$

That looks a little awkward. And we have also have to take into account that \(x\) is not discrete but a continuum. So we would also have to include states like \(| x=0.5 \rangle\) and \(| x=0.999 \rangle\). I suppose it is time to do some real mathematics. It makes more sense to replace the sum by an integral:

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

where \(\psi(x)\) is just a function of \(x\). Note also that \(| x=x \rangle \) was abbreviated to \(| x \rangle \), since it is already clear that \(x\) stands for position.

We can also ask the reverse question: how much does \(| \psi \rangle \) ‘overlap’ with \(| x=3 \rangle \) and how much with \(| x=5 \rangle\)? The answers are 0.2 and 0.9. It works much better with a formula, and we write the question “how much overlap with \(| x=3 \rangle\)?” as “\(\langle x=3|\)“, so that:

$$\langle x=3| \psi \rangle = 0.2,$$

$$\langle x=5| \psi \rangle = 0.9.$$

It is very important to remember that expressions of the form \(\langle a| b \rangle\) are just a number. In fact, we could even decompose our original superposition as:

$$| \psi \rangle = \langle x=3| \psi \rangle | x=3 \rangle + \langle x=5| \psi \rangle| x=5 \rangle.$$

Doing the same with our continuous superposition:

$$| \psi \rangle = \int_{-\infty}^{+\infty} \psi(x) |x \rangle dx = \int_{-\infty}^{+\infty} \langle x| \psi \rangle |x \rangle dx,$$

and we can also conclude that, by definition, \( \psi(x)= \langle x| \psi \rangle\).

We are now ready to reveal exactly how the velocity and the position of a particle can be combined in a single property. Firstly, it is more natural to not talk about the velocity \(v\) of the particle, but about its momentum \(p\). They are very similar, since they are related by the simple relation

$$p=mv,$$

where \(m\) is the mass of the particle. Now suppose we have a particle that has the property that it has momentum \(p\). We will write the state of the particle as \(|p\rangle\). As I told you, the position and the momentum are totally dependent properties, so we cannot add any further position information to it. This is just all there is!

What we can do, is ask how much the state \(|p\rangle\) has in common with state \(|x\rangle\). Now this is the moment when we will postulate something without deriving it, namely that that overlap is:

$$\langle x|p \rangle = e^{i x p / \hbar},$$

where \(\hbar\) is the (reduced) Planck constant and \(i^2=-1\) (see complex numbers).

So far, there really are a many aspects of quantum mechanics that I have skipped or talked very little about, but I wanted to focus on the fundamentals. It is surprising, also to me, how much we can actually already do with the extremely limited set of concepts we have developed so far. This must mean we are close to the truth! Next time, we will derive a basic model of the hydrogen atom already!