Part 7 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.

Part 6 in the quest for the hydrogen molecule

Last time we investigated the energy states of a free particle. That is of course a bit of a boring example: the particle does not really have any behavior except that it flies in one direction or another. This time we will derive our first results on how quantum mechanics leads to special bound states, which is a very important step towards understanding the hydrogen molecule.

Also we will have to roll up our sleeves, because we will have to start doing some tough mathematics do derive some real results. Do not be get discouraged, because mathematics is the language of the universe, and we need it to get to some tangible results now that we have established our first principles.

We will study the case of a single particle that is under the influence of a position-dependent potential \(V(x)\). The total energy of the particle is given by

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

We will have to choose a simple potential \(V(x)\) to work with. This is a very important choice, and for this post I would like to choose the “finite square well”,

Using the notation of Iverson brackets, the potential energy of the particle at position \(x\) is given by:

$$V(x) =-V_0[|x|<L].$$

As we did in the previous post, we will solve the energy equation to determine the energy states \(|\psi\rangle\):

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

We choose to express everything in terms of momentum states \(|p\rangle\). To make that more clear, we project both the left side and the right side on these momentum states, as was explained in part 3:

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

where we have written the projection on the momentum states as a function which we call \(\phi(p)\equiv\langle p|\psi\rangle\).

A few mathematical tools

Before we continue, let’s develop a few formulas that will be useful. First, we should also remember from part 3 that we can always project a state on the full set of either the position or the momentum states. Let me repeat those formulas here:

$$| \psi \rangle = \int_{-\infty}^{+\infty} \langle x| \psi \rangle |x \rangle dx,\\
| \psi \rangle = \int_{-\infty}^{+\infty} \langle p| \psi \rangle |p \rangle dp.$$

Note that there is an interesting case when \(| \psi \rangle = | p’ \rangle\):

$$| p’ \rangle = \int_{-\infty}^{+\infty} \langle p| p’ \rangle |p \rangle dp,$$

and since the left hand side and right hand side have to be equal, we conclude that \(\langle p| p’ \rangle =0\) whenever \(p \neq p’\). In fact, \(\langle p| p’ \rangle \) satisfies

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

for any function \(f(p)\), which will come in handy.

Solving the energy equation

Great, now we are ready to calculate. Get ready for some serious mathematics! What I would like to calculate is for a state \(|\psi\rangle\) that satisfies the energy equation:

• what energy \(E\) it might have, and
• what is its projection on the momentum states \(\phi(p)\)

First I split the kinetic and potential energy terms,

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

Kinetic energy

The first term, the kinetic energy, is relatively simple. To calculate it, I will insert a projection on momentum states \(|p’\rangle\).

$$
\langle p|\hat E_k |\psi\rangle=\langle p|\frac{\hat p^2}{2m} |\psi\rangle=\\
\int_{-\infty}^{+\infty}\! dp’\langle p|\frac{\hat p^2}{2m} |p’ \rangle \langle p’| \psi \rangle=\\
\int_{-\infty}^{+\infty}\!dp’ \langle p|\frac{p’^2}{2m} |p’ \rangle \langle p’| \psi \rangle =\\
\int_{-\infty}^{+\infty}\!dp’ \frac{p’^2}{2m} \langle p|p’ \rangle \langle p’| \psi \rangle=\\
\frac{p^2}{2m} \langle p| \psi \rangle=\frac{p^2}{2m} \phi(p).$$

Or, to summarize, the \(\hat p\) operator is also allowed to operate on the projected-to states \(\langle p|\) to the left of it.

Potential energy

Good. Now I will work on the potential term. I will use a projection on position states \(|x\rangle\):

$$\langle p|\hat E_p|\psi\rangle=\langle p|V(\hat x)|\psi\rangle=\\
\int_{-\infty}^{+\infty} \!dx\langle p|V(\hat x)\langle x| \psi \rangle |x \rangle=\\
\int_{-\infty}^{+\infty} \!dx\langle p|V(x)\langle x| \psi \rangle |x \rangle=\\
-V_0\int_{-L}^{+L} dx\langle p|x \rangle\langle x| \psi \rangle .$$

Whoa! Now it becomes clear why the square well potential is such a convenient potential! Unfortunately, this term still contains an expression \(\langle x| \psi \rangle \), while the kinetic term was expressed in terms of \(\phi(p)\). To bring the terms together again, we add another projection here, this time on momentum states \(|p’\rangle\):

$$
-V_0\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x| \psi \rangle =\\
-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x|p’ \rangle\langle p’| \psi \rangle .$$

Now we remember from part 3 our postulate:

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

which actually has a twin brother:

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

Using also by definition \(\langle p’| \psi \rangle=\phi(p’)\), we manage to arrive at a quite normal mathematical equation:

$$
\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{-i x p / \hbar}e^{i x p’ / \hbar}\phi(p’) =\\
-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar}\phi(p’) .$$

The integration over \(x\) is something we can actually do!

$$
\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar} =\\
\frac{1}{i(p’-p)}e^{i x (p’-p) / \hbar} \Biggr|_{-L}^{+L}=\\
\frac{1}{i(p’-p)}\bigg[e^{i L (p’-p) / \hbar}-e^{i L (p’-p) / \hbar}\bigg]=\\
\frac{2\sin[L (p’-p) / \hbar]}{p’-p},$$

arriving at a final expression for the potential energy:

$$\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\frac{2\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’).$$

Final equation

In the end, the full energy equation thus becomes

$$\frac{p^2}{2m}\phi(p)-2V_0\int_{-\infty}^{+\infty}\!dp’\,\frac{\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’) =E\phi(p),$$

which we have to solve for the energy value \(E\) and the function \(\phi\). How to proceed?

Numerical calculation

Unfortunately there is no analytical solution. But we can solve this equation numerically. The first thing to do is to apply some substitutions, to remove irrelevant parameters. We use.

$$
\tilde{p} \equiv \frac{L}{\hbar}p,\\
\tilde{V} \equiv \frac{2mL^2}{\hbar^2}V_0,\\
\tilde{E} \equiv \frac{2mL^2}{\hbar^2} E,\\
\tilde{\phi}(\tilde{p})\equiv \phi(p).$$

Resulting in

$$\tilde{p}^2\tilde{\phi}(\tilde{p})-2\tilde{V}\int_{-\infty}^{+\infty}\!d\tilde{p}’\,\frac{\sin(\tilde{p}’-\tilde{p})}{\tilde{p}’-\tilde{p}}\tilde{\phi}(\tilde{p}’) =\tilde{E}\tilde{\phi}(\tilde{p}) ,$$

As we can see, the solutions to the energy value \(\tilde{E}\) only depend on a single input parameter \(\tilde{V}\). Next step to solve this numerically is to replace the integral by a summation using \(\tilde{p} = n\Delta \tilde{p}\) and \(\tilde{\phi}_n=\tilde{\phi}(n\Delta \tilde{p})\):

$$(\Delta \tilde{p})^2n^2\tilde{\phi}_n-2\tilde{V}\sum_{n’=-\infty}^{+\infty}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}\tilde{\phi}_{n’} =\tilde{E}\tilde{\phi}_n,$$

which makes it look like a linear algebra problem,

$$\sum_{n’}K(n, n’)\phi_{n’}=E\phi_n,$$

with

$$K(n, n’)=((\Delta \tilde{p})^2n^2-2\tilde{V} \Delta \tilde{p})[n=n’]-2\tilde{V}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}[n \neq n’],$$

which can be solved by Octave.

Results

Free particle

First we look at the case \(\tilde{V}=0\). We expect to retrieve the energy spectrum of a free particle, since the potential well is not really there in that case. Indeed, we find the following spectrum:

All energy states have a positive energy. The reason that the energy spectrum is discrete rather than continuous is that we have replaced the integral by a summation (the distance between the states depends on the parameter \(\Delta \tilde{p}\)), but this does show nicely that the states get farther apart for higher energies.

First bound state

Moving on to a nonzero potential for the well, let’s pick \(\tilde{V}=0.5\). Here is the energy spectrum:

Remarkable! A single state was added with a negative energy. This is the energy of the bound state, and the most remarkable thing is that there is only one bound state. This does not align with our normal experience of the world. We think there are always many ways to put a marble in a shallow but wide cup: we can put it on the left or on the right, or in the middle, rolling with a little speed or not. What we learn is that this is in fact not true for very shallow cups: there is only one state, and the marble in the cup cannot carry any information.

This bound state has energy -2.1754. How does its projection on momentum states \(\phi(p)=\langle p| \psi \rangle\) and its projection on position states \(\phi(x)=\langle x| \phi \rangle\) look like?

As we can see, the bound state can be described as a mix of momentum states, or as a mix of position states. For the latter there is the unexpected discovery that the particle in the bound state does not actually stay within the limits of the well at \(x = \pm 1\). It is a strange world we live in!

Other bound states

Increasing the potential \(\tilde{V}\) lowers the energy of the first bound state, and at some point a second bound state appears. Here is a nice animation to illustrate this:

At \(\tilde{V}=3\) there are three bound states. For illustration, here are the projections on momentum states and on position states:

Conclusions

I have shown that a particle in a well has a number of discrete bound states, in contrast to the continuum of unbound states. That is an important conclusion, it is essentially where “quantum mechanics” derives its name from!

I can calculate the energy of these bound states, and also how these bound states can be written as a superposition of either momentum or position states. With respect to the momentum states, it can be noted that we have a superposition of many momentum states, and not just two of them, as one might expect. With respect to the position states, it is surprising that the particle position is not limited to within the well, although the particle in the bound state cannot escape the well.

But to me, the most remarkable thing is that we have derived all these experimentally verifiable conclusions from very few propositions. In fact we have skipped large parts of what is usually considered to be essential for understanding quantum mechanics: measurements, uncertainties and probabilities, normalization, time-evolution, the Schrödinger equation. It has been very interesting, to study this, but we are certainly not finished. In the next part, we will derive the Schrödinger equation.

Part 5 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 any states that describe 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 the kinetic energy:

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

where \(p\) is the momentum of the particle. As required, the energy of \(|\psi\rangle\) must be \(E\), we will put this down in an equation we will call the energy equation:

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

To move further, we use the fact that the energy is actually the kinetic energy, so we replace \(\hat E\) by the free-particle energy operator \(\frac{\hat p^2}{2m}\) that measures its energy in terms of momentum.

$$\frac{\hat p^2}{2m} |\psi\rangle= E |\psi\rangle.$$

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 \(|p\rangle\). 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:

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

Now we can finally get rid of the “hats” since we know that \(\hat p |p\rangle= p |p\rangle\) and therefore also \(\hat{p}^2 |p\rangle= \hat{p}\hat{p} |p\rangle= \hat{p}p |p\rangle=p\hat{p} |p\rangle= pp |p\rangle= p^2|p\rangle\).

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

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.

Part 4 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 ‘uncertainty 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:

$$| \psi \rangle = 0.2 | x=1 \rangle + 0.9| x=2 \rangle.$$

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 \(x\).

$$| \psi \rangle = 0.2 | x=1 \rangle + 0.9| x=2 \rangle+0.5| x=3 \rangle+0.4| x=4 \rangle +\ldots.$$

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 of \(| \psi \rangle \) is contributed by the \(| x=1 \rangle \) state and how much by \(| x=2 \rangle\)? The answers are 0.2 and 0.9. It works much better with a formula, and we write the question “how much overlap with \(| x=1 \rangle\)?” as “\(\langle x=1|\)“, so that:

$$\langle x=1| \psi \rangle = 0.2,$$

$$\langle x=2| \psi \rangle = 0.9.$$

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:

$$| \psi \rangle = \langle x=1| \psi \rangle | x=1 \rangle + \langle x=2| \psi \rangle| x=2\rangle.$$

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!

Part 3 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

\(|x=3\rangle\).

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.

Part 2 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

\(|\rm{happy}\rangle\).

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

\(-|\rm{happy}\rangle\).

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

\(|\rm{unhappy}\rangle\).

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

\(|\psi\rangle=|\rm{happy}\rangle+|\rm{tired}\rangle\),

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

\(|\psi\rangle=0.6|\rm{happy}\rangle+0.8|\rm{tired}\rangle\).

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

\(|\psi\rangle=|\rm{happy}\rangle+|\rm{unhappy}\rangle\).

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.

Part 1 in the quest for 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.