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Start with a system I fairly understood, the harmonic oscillator. Here all possible states fulfilling the eigenvalue-equation $H |n\rangle = E_n |n\rangle$ are given by

$$|n\rangle = \dfrac{1}{\sqrt{n!}}\,\left(\hat{b}^{+}\right)^{n} |0\rangle$$

where $\hat{b}^{+}$ represents the creation operator.

For a state in position space one has to simply multiply with a position state $\langle x|$ on both sides:

$$\langle x|n \rangle = \varphi_n = \frac{1}{\sqrt{n!}}\,\left(\hat{b}^{+}\right)^{n} \varphi_0 \, .$$

Now comes the big question: Is there an instructive analogy for the hydrogen atom? I'm aware of the relation: $\hat{L}^{+}|l,m \rangle = C_+ |l,m+1 \rangle$, but for a complete solution I would rather wish to see a representation of the form

$$|l,m \rangle = \left(A^+\right)^{l + m}|0,0 \rangle$$

or even with radial component

$$|r,l,m \rangle = \left(A^+\right)^{r + l + m}|0,0,0 \rangle$$

Does such a form exist at all, or am I requiring just nonsense?

##Extra

How to actually calculate the ground state?

Again in comparison to the harmonic oscillator, one takes use of:

$$\hat{b}|0\rangle = 0$$

where here $\hat{b}$ represents the annihilation operator. Expressing $\hat{b}$ with position coordinates one receives a differential equation for $\varphi_0$:

$$\hat{b}\varphi_0 = 0 \, .$$

To conclude: does there exists a similar equation for the hydrogen atom looking like:

$$\left({{A}}^-\right)^{r+l+m}|0,0,0\rangle = 0 ?$$

So again by expressing $A$ in position space would you get an ODE for $\varphi_0$?

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    $\begingroup$ For the hydrogen like atom, there are additional symmetries which give additional creation/annihilation like operators. Check out the Laplace-Runge-Lenz vector. However, concerning the radial part, I don’t think that an algebraic construction is possible in general for any 1D system like for the harmonic oscillator (take the particle in a box for example). $\endgroup$
    – LPZ
    Commented Feb 17, 2023 at 12:03
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    $\begingroup$ H Green, Matrix Mechanics, Ch 6.1. $\endgroup$ Commented Feb 17, 2023 at 13:00

1 Answer 1

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The Hamiltonian of the hydrogen atom, $H=\mathbf{p}^2/2m - \alpha /|\mathbf{x}|$, commutes with all components of the angular momentum operator $\mathbf{L}= \mathbf{x} \times \mathbf{p}$ and of the Laplace-Runge-Lenz-Pauli vector operator $$\mathbf{F} = \frac{1}{2}\left(\mathbf{p} \times \mathbf{L} -\mathbf{L} \times \mathbf{p}\right)- m \alpha \, \frac{\mathbf{x}}{r} =\frac{ i}{2} \, [\mathbf{p}, \mathbf{L}^2 ]-m \alpha \, \frac{\mathbf{x}}{r}, \tag{1}\label{eq1}$$ where $ r= |\mathbf{x}|$. The commutation relations $[L_k , F_\ell] = i \varepsilon_{k \ell s} F_s$ express the fact that $\mathbf{F}$ transforms as a vector under rotations. The commutator of two components of $\mathbf{F}$ can be expressed in terms of the Hamiltonian and the angular momentum, $[F_k, F_\ell ] = -2 i m H \varepsilon_{kls}L_s$. The relations $\mathbf{L} \cdot \mathbf{F} = \mathbf{F} \cdot \mathbf{L} =0$ follow immediately from $\mathbf{L} \cdot \mathbf{x} = \mathbf{L} \cdot \mathbf{p}=0$. Finally, the relation $$\mathbf{F}^2 = 2 m H (\mathbf{L}^2 +\mathbf{1})+m^2 \alpha^2 \tag{2}\label{eq2}$$ is obtained after a somewhat lengthy calculation.

Restricting ourselves on the subspace of the Hilbert space spanned by the bound states (where $H \lt 0$), the combinations $$A_k = (L_k + F_k /\sqrt{-2 m H})/2, \quad B_k =(L_k -F_k/\sqrt{-2 m H})/2 \tag{3}\label{eq3}$$ satisfy the commutation relations of two independent angular momenta, $$[A_k, A_\ell] = i \varepsilon_{k \ell s} A_s, \qquad [B_k, B_\ell ]= i \varepsilon_{k \ell s} B_s, \qquad[A_k, B_\ell ] = 0, \tag{4}\label{eq4}$$ connected by $$ \mathbf{A}^2 = \mathbf{B}^2 = -\left(\frac{1}{4}+ \frac{m \alpha^2}{8 H} \right). \tag{5}\label{eq5}$$ As we know from representation theory of the Lie algebra $\rm su(2)$, $\mathbf{A}^2$ and $\mathbf{B}^2$ have the eigenvalues $\beta (\beta+1)$ with $\beta = 0, 1/2, 1, 3/2, \ldots$, being identical for $\mathbf{A}^2$ and $\mathbf{B}^2$. Each of the eigenvectors belongs to a $(2\beta+1)^2$-degenerate supermultiplet, where the vectors can be distinguished by the eigenvalues of $(A_3, B_3)$. Using \eqref{eq5}, the eigenvalues of $H$ are given by $$ E_n = -\frac{m \alpha^2}{2 n}, \qquad n= 2 \beta +1 = 1, 2, 3, \ldots. \tag{6}\label{eq6}$$ As the eigenvalues of $\mathbf{L}^2$ are given by $\ell (\ell+1)$ ($\ell = 0,1,2, \ldots$), \eqref{eq2} implies $1-(\ell^2+\ell +1)/n \ge 0$, such that $\ell = 0, 1, \ldots n-1$ for given $n$.

In each supermultiplet, there is a state $| \, \rangle$ with maximal eigenvalues of $A_3$ and $B_3$, such that $$ A_+ | \, \rangle = B_+ | \, \rangle=0 \Leftrightarrow F_+ | \, \rangle = L_+ | \, \rangle =0 \tag{7}$$ with $A_\pm = A_1 \pm i A_2, \, B_\pm = B_1 \pm i B_2, \, F_\pm = F_1 \pm i F_2, \, L_\pm = L_1 \pm i L_2$. The other states are then obtained by applying $A_-^p B_-^q$ with $0 \le p,q \le n-1$. As $F_+$ is assembled from $x_+$, $p_+$ and terms commuting with $\mathbf{L}^2$, it raises $\ell$ by one unit, such that $F_+ | \, \rangle=0$ implies that the vector $| \, \rangle$ has already the highest possible angular momentum in this supermultiplet. In the basis $|n, \ell, \ell_3 \rangle $ of eigenvectors of $H$, $\mathbf{L}^2$, $L_3$, we have $| \, \rangle = |n, n-1, n-1\rangle $.

In the $\mathbf{x}$-representation, the equation $F_+ | n, n-1, n-1 \rangle =0$ becomes $$-\left(n\left( \frac{\partial}{\partial r}-\frac{n-1}{r}\right) + m \alpha \right) f(r) \, Y_{n-1}^{n-1}(\theta, \varphi) =0 \tag{8}$$ with the solution $$\langle \mathbf{x} | n, n-1, n-1 \rangle = c \, r^{n-1} e^{-m \alpha r/n} \, Y_{n-1}^{n-1}(\theta, \varphi). \tag{9}$$

References: Walter Thirring, Quantum Mathematical Physics, Springer, 2002, ch. 4.1; Julian Schwinger, Quantum Mechanics - Symbolism of Atomic Measurements, Springer, 2001, ch. 9.5.

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