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


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


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!

Relations between quantum states

Part 2 in the quest for the hydrogen molecule

We really have to get more exact.  The state \(|\rm{happy}\rangle\) seems a very fuzzy piece of information, there seem to be many ways to be happy, a whole continuum of them, and we require a state to describe a situation, not a property. It would be better if we could proceed with more terse and quantifiable information. We are in luck: nature absolutely loves terse and quantifiable information!

Let’s consider a particle. It has a few properties but really not that many. For instance, it has a position. Let us focus on the position in one dimension only, so that the particle has a position \(x\). That is really a well defined state. The particle can be at \(x=3\), meaning its state is


Or it can be in state \(|x=0\rangle\), \(|x=7.5\rangle\), \(|x=-4.5\rangle\) etcetera.

Remember also that all states may be added. This means that we can also have this state:

\(|x=3\rangle + |x=7\rangle\)

Meaning the superposition of the pure state of the particle at \(x=3\) and the pure state of the particle at \(x=7\). As you can see, the number of states even a single particle can attain is absolutely huge. It seems there are many more states than we would have with ‘classical’ physics, where a particle can be at only one position. However, nature turns out to be very frugal with information, as we will see, which will make a big difference. Perhaps the world of quantum mechanics is even simpler than the ‘classical’ world, but in a rather counter-intuitive way. Let’s see how this plays out.

This means we can start to build a quantum theory with a particle state, and this is how we would normally proceed. But not so quick! – you may say. If a state really corresponds to a situation, what about the situation in the rest of the universe? Should this also be part of the state? A good point, and we will see where it will lead us.

Let’s suppose the universe is made of two particles. Let’s also suppose that their position is their only property. If one of them is at \(x=3\) and the other is at \(x=5\), then the state of the universe is:

\(|x=3  \cdot x=5\rangle\).

Wait! – you might say. Should we not write down which particle is where? If particle A is at \(|x=3\rangle\) and particle B is at \(|x=5\rangle\), then perhaps the state is really \(|x_A=3 \cdot x_B=5\rangle\)? Excellent question. I have stated before that nature loves terse information, so it will not surprise you it follows the more frugal design. If the universe has only two particles, and each particle has only one property, then the complete state of the universe is described by the fact that there is a particle at \(x=3\) and there is a particle at \(x=5\), a principle with far-reaching consequences, as we will see.

If we want to do physics, we will also have to let the universe evolve over time, so let’s guess how that might work. Naturally, we expect to add a property to the particles that they have a velocity, in addition to their position. This is the state of one particle at \(x=3\) with velocity \(v=1\):

\(|x=3, v=1\rangle\).

That looks quite reasonable, but it is wrong. The velocity and the position of a particle are not independent properties. In fact, they are one and the same property! Surely, nature saw again an opportunity to save a lot of information, but how does this work precisely? We will investigate it in the next installment.

Stepping into the quantum world

Part 1 of the quest for the hydrogen molecule.

There is no way around it: we are going to do quantum mechanics. So let’s start with an introduction to establish a starting point and get ourselves familiarized with the general ideas. After that, we are going to start calculating!

There are several ways you can start building a theory of quantum mechanics. Basically these are the main options:

  1. Schrödinger equation
  2. Hilbert space states
  3. Path integrals

This is also the order by which they were discovered, and of course it also became the order in which quantum mechanics is usually taught. Following the exact footsteps of previous physicists is a great way not to make progress, since you will build the same concepts and misunderstandings in your mind as the people before you. We will try to choose our own path, where possible.

So, let’s compare them.

  1. The normal procedure would be to start with the Schröding equation and then go calculate stuff. Which is fine, but it is also not very illuminating. Where does this equation come from? What does it tell us about the world, really? It does have the advantage that you can actually calculate things, so we will have to use it at some point. It is not much use when you go to more advanced topics  (relativistic quantum mechanics, quantum field theory).
  2. Hilbert space states are nice but a a little abstract in the beginning. They are good to develop first ideas and make you feel closer to “the truth”.
  3. Path integrals are very cool, and quite useful for keeping things simple (relatively speaking) when you go to more advanced physics. They are mathematically fuzzy and strange. They are not simple to calculate and a bit frightening. We can take a look later if it is possible to calculate the hydrogen atom with path integrals, because I don’t know really.

So let’s start with the second idea, and focus on quantum states. We will learn to reason from the abstract to the concrete and we will also learn that the world is a bit more… orthogonal (?) than what we think it is.

A state is a description of a situation. We will use this notation to write down a state: \(|\rm{state}\rangle\). Furthermore the states can be added and subtracted, and they can be multiplied by any number. Don’t worry if this feels a little abstract, it’s supposed to be that way. We don’t make any assumption about the underlying structure of such a state.

Let’s start with an example. If I am feeling happy, I could describe my state as


Since states can be multiplied with -1, this is another state:


Question: which of these states is happier? Answer: they are both equally happy. An unhappy state would be an entirely different one:


What if I am both happy and tired, how can I describe that? We could write it down like this:


where I have given the combined state a name “\(|\psi\rangle\)“. The sum of two ‘pure’ states is called a superpositon. Since states can be multiplied by a number, we can also have a superposition like this


The simple rule is: all states can be added. So that means we can also have the state:


Really? It looks confusing, so we must be doing quantum mechanics!

Next time, we are going to derive the Schröding equation, although it is generally agreed that this is not possible.

The quest to understand the hydrogen molecule

I admit that I don’t really understand the hydrogen molecule.

It is the simplest molecule but it seems so complicated to have a feeling for how it really works, or to explain it well. Wouldn’t it be great to understand it thoroughly? And with it all the physical laws that make it the way it is? It would give a great view on the way things really are, to grasp the underlying simplicity that is so strange to us.

Perhaps we should go on a quest. A journey with many steps, and at each one we learn something that is real and that is true. It will be a difficult quest. We may not even make it. Not all of us will make it. And we may take wrong turns or have to come back to a point where we were before. We may take more difficult steps where we could have taken a simpler path. Who knows where we will end up. But it will be quite educational. If you can follow the path. We’ll see.

Save money by driving

Do you enjoy driving? Do you care about pollution? Do you like money? If your answer to one of these questions is yes, the following may be of interest to you.

Last summer it was quiet on the road during my daily commute. This enabled me to do a little experiment that I have been wanting to do. I know my car uses less fuel when driving slower, but how much less?

To test this, every day I would drive 100 km/h on the highway in the morning and 120 km/h in the evening when coming back, or vice versa, chosen more or less randomly. Every time I recorded fuel consumption for my Fiat 500, travel time, and temperature.

This gave me some data points, that I could analyze. Now to be fair, I have only 13 data points, so it is not all very scientific. But citizen science is mostly about the fun, right?

Here are the data points I recorded, plotted against temperature:

Fuel consumption versus temperature

Now be careful, the fuel consumption is for the whole trip, while the change in driving speed was only applied to the highway part. Let’s see if the trip time correlates with the speed.

Time vs inverse speed

              Estimate Std. Error t value Pr(>|t|)    
(Intercept)     21.143      3.391   6.235 6.39e-05 ***
inverse_speed   40.714      6.098   6.677 3.48e-05 ***

Residual standard error: 1.096 on 11 degrees of freedom
Multiple R-squared:  0.8021,    Adjusted R-squared:  0.7841 
F-statistic: 44.58 on 1 and 11 DF,  p-value: 3.483e-05

(sorry for being confusing with the inverse speed, but it is the physically correct quantity to use). That correlation is certainly significant, but there is some variation. Probably this is due to random events, like traffic lights or small traffic jams. The correlation is 40.714 km, meaning that effectively 40.7 km of the 60 km drive was on highways.

The question rises whether it is better to use the correlation of fuel consumption with speed, or with travel time. Because of the variations mentioned above, I actually find a more significant correlation with speed. I will use this to calculate how much money you save when driving slower.

Here are the correlations with all parameters that seem relevant:

             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.055750   0.655437   3.136 0.011996 *  
speed        0.028449   0.004505   6.316 0.000138 ***
temperature -0.028659   0.027249  -1.052 0.320333    
direction    0.290119   0.144276   2.011 0.075223 .  

Residual standard error: 0.1496 on 9 degrees of freedom
Multiple R-squared:  0.8734,    Adjusted R-squared:  0.8312 
F-statistic:  20.7 on 3 and 9 DF,  p-value: 0.0002236

All correlations for fuel consumption

Indeed, the correlation with speed is very significant. There is also a correlation with direction (morning or afternoon drive), probably due to typical wind conditions. The correlation with temperature is not very significant, but in my experience it really becomes significant when you compare summer and winter driving conditions.

The speed / fuel consumption coefficient is 0.028449, or, in other words, by driving 20 km/h slower I save 0.56898 liter per 100 km. But I assumed that the savings were only due to the highway part of our trip. So I should calculate the absolute numbers: the full trip is 59.9 km, so the saving is 0.3408 liter. For just the highway part, this is 0.8374 liter per 100 km.

The average price of Euro95 right now is 1.600 euro/liter. This means that I save 1.34 euro per 100 km. When driving 120 km/h, driving 100 km takes 50 minutes, but at 100 km/h, it takes 60 minutes, a difference of 10 minutes.

Of course, I assume here that all the correlations are linear, but I know that they are not. In fact, if you drive faster, say 130 km/h or 140 km/h, your fuel consumption will increase more dramatically, and the savings will be bigger.


By driving slower on the highway, you can save 1.34 euro, but you arrive 10 minutes later. This means you save 8 euro by driving for one hour. Maybe not enough to earn a living, but if you have the time: relax and drive slower!

If you are interested, you can get the  data and the R script that I used.