The hydrogen energy eigenstates are labelled with well-defined $n$, the principal quantum number. Is there an associated operator than one can apply to these wavefunctions which would result in an eigenvalue of $n$ multiplied by the hydrogen wavefunction? If so, what does the operator look like?


2 Answers 2


You can always construct an operator that gives you any quantum number you wish, using the eigenstates. Let the bound states of the hydrogen atom be $|n,l,m\rangle$; the representations of these states in position space (that is, the spatial wave functions) are well known. Then you may have an operator $${\cal N}=\sum_{n=1}^{\infty}\sum_{l=0}^{n-1}\sum_{m=-l}^{l}n|n,l,m\rangle\langle n,l,m|,$$ for which the eigenvalues are the principal quantum numbers for the bound states and zero otherwise.

In the position space representation of this operator, each term in the sum involves the explicit Schrödinger wave function. For example, when the $n=1$ term acts on a general wave function $\Psi(\vec{r})$, it takes the form $$\psi_{1,0,0}(\vec{r})\int d^{3}r'\,\psi^{\dagger}_{1,0,0}\left(\vec{r}\,'\right)\Psi\left(\vec{r}\,'\right) =\frac{1}{\pi a_{0}^{3}}\exp\left(-r/a_{0}\right)\int d^{3}r'\, \exp\left(-r'/a_{0}\right)\Psi\left(\vec{r}\,'\right).$$ You could also write this operator in a different basis of energy eigenstates (such as that obtained by solving the Coulomb problem in parabolic coordinates).

However, if you asking whether this has a convenient form when expressed in terms of $\vec{r}$ and $\vec{p}$, the answer is no. For one thing, the principal quantum number is not something that is even defined for the complete hydrogen spectrum; the continuum states above zero energy do not have a value for $n$, and if you tried to continue $n$ to these unbound states, its values would be imaginary. If you apply the explicit operator ${\cal N}$ to a continuum energy eigenstate, you will get $0$, since the bound states and continuum states are orthogonal.


The principal quantum number $n$ corresponds to the energy of the state $|n\ell m\rangle$. In other words, the operator associated to $n$ is the Hamiltonian $H$, and, for the hydrogen atom,

$$ H|n\ell m\rangle = E_n|n\ell m\rangle,$$

where $E_n = -E_I / n^2$, and $E_I$ is the ionization energy of the hydrogen atom.


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