Ladder operator

Many terms in the Hamiltonians of atomic or molecular systems involve the scalar product of angular momentum operators. An example is the magnetic dipole term in the hyperfine Hamiltonian,[4]

$${\displaystyle {\hat {H}}_{\text{D}}={\hat {A}}\mathbf {I} \cdot \mathbf {J} ,}$$

where I is the nuclear spin.

The angular momentum algebra can often be simplified by recasting it in the spherical basis. Using the notation of spherical tensor operators, the "-1", "0" and "+1" components of J⁽¹⁾ ≡ J are given by,[5]

$${\displaystyle {\begin{aligned}J_{-1}^{(1)}&={\dfrac {1}{\sqrt {2}}}(J_{x}-iJ_{y})={\dfrac {J_{-}}{\sqrt {2}}}\\J_{0}^{(1)}&=J_{z}\\J_{+1}^{(1)}&=-{\frac {1}{\sqrt {2}}}(J_{x}+iJ_{y})=-{\frac {J_{+}}{\sqrt {2}}}.\end{aligned}}}$$

From these definitions, it can be shown that the above scalar product can be expanded as

$${\displaystyle \mathbf {I} ^{(1)}\cdot \mathbf {J} ^{(1)}=\sum _{n=-1}^{+1}(-1)^{n}I_{n}^{(1)}J_{-n}^{(1)}=I_{0}^{(1)}J_{0}^{(1)}-I_{-1}^{(1)}J_{+1}^{(1)}-I_{+1}^{(1)}J_{-1}^{(1)},}$$

The significance of this expansion is that it clearly indicates which states are coupled by this term in the Hamiltonian, that is those with quantum numbers differing by m_i = ±1 and m_j = ∓1 only.

Another application of the ladder operator concept is found in the quantum mechanical treatment of the electronic energy of hydrogen-like atoms and ions. The Laplace–Runge–Lenz vector commutes with the Hamiltonian for an inverse square spherically symmetric potential and can be used to determine ladder operators for this potential.[6][7] We can define the lowering and raising operators (based on the classical Laplace–Runge–Lenz vector)

$${\displaystyle {\vec {A}}=\left({\frac {1}{Ze^{2}\mu }}\right)\left\{{\vec {L}}\times {\vec {p}}-{\boldsymbol {i}}\hbar {\vec {p}}\right\}+{\frac {\vec {r}}{r}}}$$

where ${\displaystyle {\vec {L}}}$ is the angular momentum, ${\displaystyle {\vec {p}}}$ is the linear momentum, ${\displaystyle \mu }$ is the reduced mass of the system, ${\displaystyle e}$ is the electronic charge, and ${\displaystyle Z}$ is the atomic number of the nucleus. Analogous to the angular momentum ladder operators, one has ${\displaystyle A_{+}=A_{x}+iA_{y}}$ and ${\displaystyle A_{-}=A_{x}-iA_{y}}$.

The commutators needed to proceed are:

$${\displaystyle [A_{\pm },L_{z}]=\mp {\boldsymbol {i}}\hbar A_{\mp }}$$

and

$${\displaystyle [A_{\pm },L^{2}]=\mp 2\hbar ^{2}A_{\pm }-2\hbar A_{\pm }L_{z}\pm 2\hbar A_{z}L_{\pm }.}$$

Therefore,

$${\displaystyle A_{+}|?,\ell ,m_{\ell }\rangle \rightarrow |?,\ell ,m_{\ell }+1\rangle }$$

and

$${\displaystyle -L^{2}\left(A_{+}|?,\ell ,\ell \rangle \right)=-\hbar ^{2}(\ell +1)((\ell +1)+1)\left(A_{+}|?,\ell ,\ell \rangle \right)}$$

so

$${\displaystyle A_{+}|?,\ell ,\ell \rangle \rightarrow |?,\ell +1,\ell +1\rangle }$$

where the "?" indicates a nascent quantum number which emerges from the discussion.

Given the Pauli[8][9] equations Pauli Equation IV:

$${\displaystyle 1-A\cdot A=-\left({\frac {2E}{\mu Z^{2}e^{4}}}\right)(L^{2}+\hbar ^{2})}$$

and Pauli Equation III:

$${\displaystyle \left(A\times A\right)_{j}=-\left({\frac {2{\boldsymbol {i}}\hbar E}{\mu Z^{2}e^{4}}}\right)L_{j}}$$

and starting with the equation

$${\displaystyle A_{-}A_{+}|\ell ^{*},\ell ^{*}\rangle =0}$$

and expanding, one obtains (assuming ${\displaystyle \ell ^{*}}$ is the maximum value of the angular momentum quantum number consonant with all other conditions),

$${\displaystyle \left(1+{\frac {2E}{\mu Z^{2}e^{4}}}(L^{2}+\hbar ^{2})-i{\frac {2i\hbar E}{\mu Z^{2}e^{4}}}L_{z}\right)|?,\ell ^{*},\ell ^{*}\rangle =0}$$

which leads to the Rydberg formula:

$${\displaystyle E_{n}=-{\frac {\mu Z^{2}e^{4}}{2\hbar ^{2}(\ell ^{*}+1)^{2}}}}$$

implying that ${\displaystyle \ell ^{*}+1=n=?}$, where ${\displaystyle n}$ is the traditional quantum number.

The Hamiltonian for a hydrogen-like potential can be written in spherical coordinates as

$${\displaystyle H={\frac {1}{2\mu }}\left[p_{r}^{2}+{\frac {1}{r^{2}}}L^{2}\right]+V(r)}$$

where ${\displaystyle V(r)=-Ze^{2}/r}$ and ${\displaystyle p_{r}}$ is the radial momentum

$${\displaystyle p_{r}={\frac {x}{r}}p_{x}+{\frac {y}{r}}p_{y}+{\frac {z}{r}}p_{z}}$$

which is real and self-conjugate.

Suppose ${\displaystyle |nl\rangle }$ is an eigenvector of the Hamiltonian where ${\displaystyle l\rangle }$ is the angular momentum and ${\displaystyle n}$ represents the energy, so ${\displaystyle L^{2}|nl\rangle =l(l+1)\hbar ^{2}|nl\rangle }$ and we may label the Hamiltonian as ${\displaystyle H_{l}}$

$${\displaystyle H={\frac {1}{2\mu }}\left[p_{r}^{2}+{\frac {1}{r^{2}}}l(l+1)\hbar ^{2}\right]+V(r)}$$

The factorization method was developed by Infeld and Hull[10] for differential equations. Newmarch and Golding[11] applied it to spherically symmetric potentials using operator notation.

Suppose we can find a factorization of the Hamiltonian by operators ${\displaystyle C_{l}}$ as

${\displaystyle C_{l}^{*}C_{l}=2\mu H_{l}+F_{l}}$

(1)

and

$${\displaystyle C_{l}C_{l}^{*}=2\mu H_{l+1}+G_{l}}$$

for scalars ${\displaystyle F_{l}}$ and ${\displaystyle G_{l}}$. The vector ${\displaystyle C_{l}C_{l}^{*}C_{l}|nl\rangle }$ may be evaluated in two different ways as

$${\displaystyle {\begin{aligned}C_{l}C_{l}^{*}C_{l}|nl\rangle &=(2\mu E_{l}^{n}+F_{l})C_{l}|nl\rangle \\&=(2\mu H_{l+1}+G_{l})C_{l}|nl\rangle \end{aligned}}}$$

which can be re-arranged as

$${\displaystyle H_{l+1}(C_{l}|nl\rangle )=[E_{l}^{n}+(F_{l}-G_{l})/(2\mu )](C_{l}|nl\rangle )}$$

showing that ${\displaystyle C_{l}|nl\rangle }$ is an eigenstate of ${\displaystyle H_{l+1}}$ with eigenvalue

$${\displaystyle E_{l+1}^{n^{'}}=E_{l}^{n}+(F_{l}-G_{l})/(2\mu )}$$

If ${\displaystyle F_{l}=G_{l}}$ then ${\displaystyle n^{'}=n}$ and the states ${\displaystyle |nl\rangle }$ and ${\displaystyle C_{l}|nl\rangle }$ have the same energy.

For the hydrogenic atom, setting

$${\displaystyle V(r)=-{\frac {B\hbar }{\mu r}}}$$

with

$${\displaystyle B={\frac {Z\mu e^{2}}{\hbar }}}$$

a suitable equation for ${\displaystyle C_{l}}$ is

$${\displaystyle C_{l}=p_{r}+{\frac {i\hbar (l+1)}{r}}-{\frac {iB}{l+1}}}$$

with

$${\displaystyle F_{l}=G_{l}={\frac {B^{2}}{(l+1)^{2}}}}$$

There is an upper bound to the ladder operator if the energy is negative, (so ${\displaystyle C_{l}|nl_{\max }\rangle =0}$ for some ${\displaystyle l_{\max }}$) then from Equation (1)

$${\displaystyle E_{l}^{n}=-F_{l}/{2\mu }=-{\frac {B^{2}}{2\mu (l_{\max }+1)^{2}}}=-{\frac {\mu Z^{2}e^{4}}{2\hbar ^{2}(l_{\max }+1)^{2}}}}$$

and ${\displaystyle n}$ can be identified with ${\displaystyle l_{\max }+1}$

Whenever there is degeneracy in a system, there is usually a related symmetry property and group. The degeneracy of the energy levels for the same value of ${\displaystyle n}$ but different angular momenta has been identified as the SO(4) symmetry of the spherically symmetric Coulomb potential.[12][13]

A suitable factorization is given by[11]

$${\displaystyle C_{l}=p_{r}+{\frac {i\hbar (l+1)}{r}}-i\mu \omega r}$$

with

$${\displaystyle F_{l}=-(2l+3)\mu \omega \hbar }$$

and

$${\displaystyle G_{l}=-(2l+1)\mu \omega \hbar }$$

Then

$${\displaystyle E_{l+1}^{n^{'}}=E_{l}^{n}+{\frac {F_{l}-G_{l}}{2\mu }}=E_{l}^{n}-\omega \hbar }$$

and continuing this,

$${\displaystyle {\begin{aligned}E_{l+2}^{n^{'}}&=E_{l}^{n}-2\omega \hbar \\E_{l+3}^{n^{'}}&=E_{l}^{n}-3\omega \hbar \\&\;\;\vdots \end{aligned}}}$$

Now the Hamiltonian only has positive energy levels as can be seen from

$${\displaystyle {\begin{aligned}\langle \psi |2\mu H_{l}|\psi \rangle &=\langle \psi |C_{l}^{*}C_{l}|\psi \rangle +\langle \psi |(2l+3)\mu \omega \hbar |\psi \rangle \\&=\langle C_{l}\psi |C_{l}\psi \rangle +(2l+3)\mu \omega \hbar \langle \psi |\psi \rangle \\&\geq 0\end{aligned}}}$$

This means that for some value of ${\displaystyle l}$ the series must terminate with ${\displaystyle C_{l_{\max }}|nl_{\max }\rangle =0}$ and then

$${\displaystyle E_{l_{\max }}^{n}=-{\frac {F_{l_{\max }}}{2\mu }}=\left(l_{\max }+{\frac {3}{2}}\right)\omega \hbar }$$

This is decreasing in energy by ${\displaystyle \omega \hbar }$ unless for some value of ${\displaystyle l}$, ${\displaystyle C_{l}|n,l\rangle =0}$. Identifying this value as ${\displaystyle n}$ gives

$${\displaystyle E_{l}^{n}=-F_{l}=\left(n+{\tfrac {3}{2}}\right)\omega \hbar }$$

It then follows the ${\displaystyle n'=n-1}$ so that

$${\displaystyle C_{l}|nl\rangle =\lambda _{l}^{n}|n-1,\,l+1\rangle }$$

giving a recursion relation on ${\displaystyle \lambda }$ with solution

$${\displaystyle \lambda _{l}^{n}=-\mu \omega \hbar {\sqrt {2(n-l)}}}$$

There is degeneracy caused from angular momentum; there is additional degeneracy caused by the oscillator potential. Consider the states ${\displaystyle |n,\,n\rangle ,|n-1,\,n-1\rangle ,|n-2,\,n-2\rangle ,\dots }$ and apply the lowering operators ${\displaystyle C^{*}}$: ${\displaystyle C_{n-2}^{*}|n-1,\,n-1\rangle ,C_{n-4}^{*}C_{n-3}^{*}|n-2,\,n-2\rangle ,\dots }$ giving the sequence ${\displaystyle |n,n\rangle ,|n,\,n-2\rangle ,|n,\,n-4\rangle ,\dots }$ with the same energy but with ${\displaystyle l}$ decreasing by 2. In addition to the angular momentum degeneracy, this gives a total degeneracy of ${\displaystyle (n+1)(n+2)/2}$[14]

The degeneracies of the 3D isotropic harmonic oscillator are related to the special unitary group SU(3)[14][15]