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When solving the Schrödinger equation for the hydrogen atom, textbooks invariably work in a more constraint situation, whereby not only an eigenfunction for the Hamiltonian operator $\hat{H}$ is sought, but one which is simultaneously an eigenfunction for $\hat{L}^2$ and $\hat{L}_z$. My question is why we do this?

A similar question has been asked here, but the answers are unsatisfactory. Yes, I understand we can do it. Yes, I understand that we have lots of freedom in our choice of $\psi(\vec{x})$ if we merely solve for $\hat{H}$. But I want to know why this is the right way to proceed. As far as I understand, it is perfectly physically acceptable for a wave function to not be an eigenfunction of some operator, so why must the wave function for a hydrogen atom be an eigenfunction for $\hat{L}^2$ and $\hat{L}_z$?

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    $\begingroup$ The physical state of a hydrogen atom doesn't have to be an eigenstate of all these operators. But it can always be written as a linear combination of those eigenstates. In quantum mechanics, we're always looking for a basis of the Hilbert space which makes calculations convenient. The set of eigenstates of the given operators is such a convenient basis. But there's nothing preventing you from finding an atom in a state which is not one of those eigenstates. It will simply take more symbols to write it down, for instance $\frac{1}{\sqrt2}(\vert n=1\rangle+\vert n=2\rangle)$. $\endgroup$ – Vercassivelaunos Aug 1 '20 at 10:08
  • $\begingroup$ Thanks @Vercassivelaunos. I was under the impression that this would only be true when considering eigenvalues of a single operator, but it seems that I am mistaken. $\endgroup$ – Mr. Palomar Aug 1 '20 at 10:12
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I'd like to elaborate some more on my comment. As you probably know, quantum mechanics take place in a Hilbert space $\mathcal H$, and every physical state is represented by a unit vector in $\mathcal H$. The time evolution of a given state $\vert\psi\rangle$ is given by the Schrödinger equation $i\hbar\vert\dot\psi\rangle=H\vert\psi\rangle$. Using some math we can show that if $\vert\psi(t=0)\rangle$ is an eigenstate of the Hamiltonian $H$, solving this equation becomes particularly easy (just multiply $\vert\psi(t=0)\rangle$ by an appropriate phase factor). And if we can express any arbitrary state as a linear combination of eigenstates, it's still pretty easy (give each term a phase factor of its own). So solving the Schrödinger equation reduces to finding a basis of the Hilbert space made up entirely of eigenstates of the Hamiltonian (because then we can express any arbitrary state as a linear combination of eigenstates, and thus solve the equation as described above).

Now the main part of a lecture about the hydrogen atom will consist of finding this basis in different scenarios (with or without spin, with strong/weak/no electromagnetic fields, etc.). And it turns out that the Hamiltonian of the hydrogen atom is degenerate, so we have some free choices when looking for a basis. And it turns out that very conveniently, we have the choice to make the basis states into eigenstates of not only the Hamiltonian, but also additional physically relevant operators: $L^2,L_z,S^2$ and $S_z$.

This does not mean that the basis basis states are the only physically allowed states. Just that all physically allowed states can be expressed as a linear combination of these eigenstates. It's actually quite unlikely to find an atom in any of those basis eigenstates. For instance, the electron might have a defined energy quantum number $n=2$, defined absolute angular momentum number $l=1$, but the direction of the angular momentum might not be along our arbitrarily chosen $z$-axis, so the electron has no defined quantum number $m_l$. So maybe it's in the state $\vert\psi\rangle=\frac{1}{\sqrt2}(\vert n=2,l=1,m_l=1\rangle+\vert n=2,l=1,m_l=-1\rangle)$, which is not one of our basis states. But we can still use the time evolution of the basis states to calculate the time evolution of this other state, too.

There are also different choices for a basis, and every physically possible state can still be written as a linear combination of those. But the one that is found in the standard literature is the most convenient when trying to solve the Schrödinger equation.

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  • $\begingroup$ Thanks, I think I'm starting to understand better. Is this situation unique for the hydrogen atom? E.g. if we take an atom with more electrons, is there still a finite set of operators whose eigenvalues uniquely specify a basis of $\mathcal{H}$? (I am aware that the Schrödinger equation for atoms with more electrons cannot be solved in terms of exact formulas, but that doesn't exclude such a basis from existing, I presume.) $\endgroup$ – Mr. Palomar Aug 1 '20 at 11:21
  • $\begingroup$ You can always find a basis of the Hilbert space made up of eigenstates of the Hamiltonian, and you can then just invent a fitting set of operators by specifying how they act on these eigenstates, so yes. Wether they are physically relevant is a different question, though. $\endgroup$ – Vercassivelaunos Aug 1 '20 at 12:32
  • $\begingroup$ Thanks, that makes sense. $\endgroup$ – Mr. Palomar Aug 1 '20 at 13:12
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The question that you have shared does deal with this point, but I thought I'd elaborate on it a little more. When we solve a quantum mechanical system, we would like to work with an eigenbasis that is labelled by quantum numbers that completely specify the state of the system. Essentially, this comes down to finding something called a Complete Set of Commuting Observables.

The wikipedia article linked gives a good introduction, but basically a "CSCO" is the largest set of operators that can be formed such that we can speak of the eigenvalues of all of these operators simultaneously. (For example, $\hat{x}$ and $\hat{p}$ would never be in a CSCO since there exists no eigenbasis that is simultaneously an eigenbasis of $\hat{x}$ and $\hat{p}$.)

In the case of the hydrogen atom, there is a large amount of degeneracy. Solving the Schrodinger equation, it can be shown that for a state of definite energy $n$, the degeneracy is $n^2$. (I'll come back to this further down.) But all of these $n^2$ states are not equivalent. While they all do have the same energy, they differ in their values of total and azimuthal angular momentum. Finding an eigenbasis that is a simultaneous eigenbasis of all of these operators would allow us to label each of these degenerate states with two more numbers, which makes them "unique".

But, you may ask, how do we know that these three operators are sufficient to form the CSCO? To my understanding, there's no way for us to know, apart from experiment. It turns out that $\{\hat{H}, \hat{L}^2, \hat{L}_z\}$ don't form a CSCO on their own! To completely specify a state of an electron in the hydrogen atom one is also required to specify the Spin of the electron, and so the true CSCO is $\{\hat{H}, \hat{L}^2, \hat{L}_z, \hat{S}_z\}$. (This is why the true degeneracy of the hydrogen atom is $2n^2$, as there are two values of spin possible per state.)

So as I see it, the states of the Hydrogen atom don't have to be eigenstates of all three operators. But if we would like each physically distinguishable state to be represented by a unique vector in the Hilbert Space, then we need to identify each eigenvector by the largest set of eigenvalues that uniquely specifies it. Ignoring spin, these eigenvalues are $|n l m\rangle$.

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  • $\begingroup$ Thanks for the response. The question whether a set of operators forms a CSCO isn't merely a mathematical one, correct? For instance, when adding spin, what we're doing is we enlarge the ambient Hilbert space (tensoring with $\mathbb{C}^2$, I guess) so that there's more 'space' for an additional operator. Had we not enlarged it in the first place, $\big\{\hat{H},\hat{L}^2,\hat{L}_z\big\}$ would've been a perfectly fine CSCO. Do I understand that correctly? $\endgroup$ – Mr. Palomar Aug 1 '20 at 11:24
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    $\begingroup$ @Mr.Palomar I'm not completely sure I understand, but yes to the first part. The question as to what determines a CSCO isn't (as far as I know) one that can be solved entirely mathematically. If we had "ignored" spin and thought that a system could be completely specified by the eigenvalues $| nlm\rangle$, then we'd be surprised by what happens when we place the atom in an external magnetic field: we would get more splittings in the spectral lines than expected! (Indeed this is why the "true" Zeeman effect is called "anomalous": it couldn't be explained without introducing Spin.) $\endgroup$ – Philip Aug 1 '20 at 11:43
  • $\begingroup$ Thanks for the response, it makes sense to me now. $\endgroup$ – Mr. Palomar Aug 1 '20 at 13:13
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It is very desirable for $\psi$ to be an eigenfunction of as many operators as possible

In fact, we would probably like it to be an eigenfunction of all angular momenta, but they do not commute, so we can't make it be an eigenfunction of all momenta, so we choose only one (usually $L_z$)

There are many reasons, but the main ones are:

  1. They are measurable

The Hamiltonian is the energy operator (roughly speaking). If a function is an eigenstate of the Hamiltonian, that means that $\phi_n$ has a defined energy. Energy is a quantity which is easily measurable, so we choose the Hamiltonian to be an important operator in our complete system of commuting operators (CSCO)

In the same way, angular momenta are easily measurable using magnetic fields.

  1. The fact that a function is an eigenstate of an operator allows us to "label" it with quantum numbers. We can set a state to be $|n\ l\ m_l\ s \ m_s\rangle$ because of that. You cannot have a quantum number if the function is not an eigenstate, because it wouldn't have an eigenvalue to label it with.

So having "something easily measurable to label with" is a good idea. Saying $n=1$ is good because it is easily measurable.

Plus, it is a really basic concept in physics, and we have a lot of intuition on it. If we say that the energy level is the first level, we quickly get an idea of how the electron is. This would not happen with weirder magnitudes.

  1. The Hamiltonian is more than that

The Hamiltonian, besides the energy, is also responsible of the time evoluition. As it is involved in the Schrödinger's equation, the Hamiltonian rules the time evolution. If something commutes with the Hamiltonian, then that quantity is conserved over time. So if $[H, L_z]=O$, then the value you measure for $L_z$ is conserved over time. So a state with $m_s=+1$, for example, will keep that value over time. That's why the Hamiltonian is important.

The angular momenta are important as well, because things that commute with the angular momentum are invariant under rotations, which is useful, because it tells us if things are symmetric or not. You know that symmetry plays an important role when simplifying problems.

  1. Commutation itself

The very fact that two operators commute is probably always a good thing. Commutation means that you can measure $H$ and then $L_z$ and vice-versa, and the result is the same. Translation: if you measure energy, it doesn't disrupt the system and you can measure $L_z$ afterwards. If they do not commute, measuring one changes the state for the next measurement.

So, you want them to commute, because you want to measure both things of the same state. It happens to be that "knowing" $L_z$ is not enough to determine the state, there are many states with the same value of $m_l$. We need more information.

However, given a certain energy, with a certain momentum, and certain spin, and so on, the state is unambiguously determined.

That's why we need a complete set of operators. And they must commute so that we can measure all of them without disrupting the system under measurement. That's why we seek for a CSCO.

And, of course, if you need 5 operators to determine your system, you'd better choose operators whose magitudes are easily measurable and have a good physical meaning.

And there are probably more reasons, but I can't remember them all now. Feel free to complete my answer in comments.

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  • $\begingroup$ Thanks for the response. Many of the motivations for doing so, however, seem to be from the point of view of the physicist. We benefit from symmetry, we benefit from getting quantum numbers, etc. But the system itself need not listen to our hopes per se. Why does the system have a particular tendency to be simultaneous eigenfunctions? $\endgroup$ – Mr. Palomar Aug 1 '20 at 10:07
  • $\begingroup$ @Mr.Palomar Sorry, I don't understand what you mean $\endgroup$ – FGSUZ Aug 1 '20 at 11:04
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" so why must the wave function for a hydrogen atom be an eigenfunction for $\hat{L}^2$ and $\hat{L}_z$?" They must not, but since the $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ do commute with each other, they can have some mutual eigenfunctions.You do not have to choose these specific mutual eigenfunctions, you can choose some others, but I think it is obviously very convenient to work with such functions that are simultaneously eigenfunctions of all these three operators(and this is why people usually choose these ones).

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  • $\begingroup$ A car can, but must not, travel 50 mph. To assume that all cars in the universe drive 50 mph makes your analysis more convenient, but the results it's going to yield are physically insignificant. So, too, is forcing my wave function to be a simultaneous eigenfunction very convenient, but unless there is a reason why the constraint is logical, the outcomes are physically unmotivated. $\endgroup$ – Mr. Palomar Aug 1 '20 at 9:56
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    $\begingroup$ Actually the whole thing about resolving physical problems is to find "the most convenient way/instrument". One does not have to solve the problem of cannon shooting in the Earth's frame of reference(one can use the sun for instance as a frame of reference ), but if one is sane he uses the Earth's frame of reference. $\endgroup$ – Leiba Goldstein Aug 1 '20 at 10:40
  • $\begingroup$ But then the outcome will always be the same. I mistakenly thought that this is not the case in the situation of the hydrogen atom, in that the assumption of being an eigenfunction of $\hat{L}^2$ and $\hat{L}_z$ has explicit influence on the outcome. Namely, you find some functions $R(r)\,Y^{m}_{\ell}(\theta,\phi)$, and I figured that every wave function must always be of that form, while this wouldn't be the case had we not made the assumption about being an eigenfunction of $\hat{L}^2$ and $\hat{L}_z$. But I forgot that superposition was a thing; we merely chose a good basis to work in. $\endgroup$ – Mr. Palomar Aug 1 '20 at 11:11
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The nuclear potential is rotationally symmetric so all eigenfunctions must be simultaneously eigenfunctions of angular momentum.

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    $\begingroup$ "all eigenfunctions must be simultaneously eigenfunctions of angular momentum". This is incorrect. They do not are necessarily simultaneous eigenfunctions. Rather, there exists a basis of such simultaneous eigenfunctions, which is particularly convenient. But that is not the only possible basis made of energy eigenstates. $\endgroup$ – HicHaecHoc Aug 1 '20 at 10:39
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    $\begingroup$ Counterexample: take any two states with same quantum number $n$ but different quantum number $l$. Their linear combination will still be an energy eigenstate , but not an angular momentum eigenstate. $\endgroup$ – Vercassivelaunos Aug 1 '20 at 10:42

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