Angular momentum operator

Angular momentum operator is a term used in quantum mechanics . It is a Hermitian vector operator whose components satisfy the following commutator relation: ${\displaystyle {\vec {J}}=({J}_{x},{J}_{y},{J}_{z})}$

${\displaystyle [{J}_{a},{J}_{b}]=\mathrm {i} \hbar \varepsilon _{abc}{J}_{c}}$

${\displaystyle \varepsilon _{abc}}$is the Levi Civita symbol . The square of the angular momentum operator is defined as the sum of the squares of its components:

${\displaystyle {\vec {J}}^{2}={J}_{x}^{2}+{J}_{y}^{2}+{J}_{z}^{2}}$.

The angular momentum operator plays a central role in understanding atoms and other quantum mechanical problems with rotational symmetry . The orbital angular momentum operator is the quantum mechanical analogue of the classical angular momentum . There is also the spin operator , which is also an angular momentum operator, but does not have a classic analog. ${\displaystyle {\vec {L}}}$ ${\displaystyle {\vec {S}}}$

properties

In the following, Einstein's summation convention is used, which means that double indices are added together.

From the commutator relation it follows automatically: ${\displaystyle [{J}_{a},{J}_{b}]=\mathrm {i} \hbar \varepsilon _{abc}{J}_{c}}$

${\displaystyle [{J}_{a},{\vec {J}}^{2}]=0}$

Since the components of the angular momentum operator do not interchange by definition, one often goes over to the common eigenvectors of and any angular momentum component (usually ). denote the eigenvectors of the common basis of and and the following eigenvalue equations apply : ${\displaystyle {\vec {J}}^{2}}$ ${\displaystyle {J}_{z}}$ ${\displaystyle |jm\rangle }$ ${\displaystyle {\vec {J}}^{2}}$ ${\displaystyle {J}_{z}}$

${\displaystyle {\vec {J}}^{2}|jm\rangle =\hbar ^{2}j(j+1)|jm\rangle }$

${\displaystyle {J}_{z}|jm\rangle =\hbar m|jm\rangle }$

The quantum number can assume the values and the quantum number the values . Thus, the same fold degenerate . ${\displaystyle j}$ ${\displaystyle j={\frac {n}{2}},\ n\in \mathbb {N} _{0}}$ ${\displaystyle m}$ ${\displaystyle m=-j,-j+1,\dots ,j}$ ${\displaystyle j}$ ${\displaystyle (2j+1)}$

In the case of orbital angular momentum, the indices and correspond to the secondary quantum number ( integer) or the magnetic quantum number of the orbital angular momentum and, similarly, in the case of spin, to the two spin quantum numbers ( half- integer ) and . ${\displaystyle j}$ ${\displaystyle m}$ ${\displaystyle {\vec {L}}}$ ${\displaystyle l}$ ${\displaystyle l}$ ${\displaystyle m}$ ${\displaystyle {\vec {S}}}$ ${\displaystyle s}$ ${\displaystyle s}$ ${\displaystyle m_{s}}$

One defines ladder operators with which the spectrum can be traversed given : ${\displaystyle m}$ ${\displaystyle j}$

• Ascent operator: ${\displaystyle {J}_{+}={J}_{x}+\mathrm {i} {J}_{y}}$

${\displaystyle {J}_{+}|j,m\rangle =\hbar {\sqrt {(j-m)(j+m+1)}}|j,m+1\rangle =\hbar {\sqrt {j(j+1)-m(m+1)}}|j,m+1\rangle }$

• Relegation operator: ${\displaystyle {J}_{-}={J}_{x}-\mathrm {i} {J}_{y}}$

${\displaystyle {J}_{-}|j,m\rangle =\hbar {\sqrt {(j+m)(j-m+1)}}|j,m-1\rangle =\hbar {\sqrt {j(j+1)-m(m-1)}}|j,m-1\rangle }$

Details on the quantum numbers j and m

The possible eigenvalues and are determined from and using ladder operators . First, different commutators with ladder operators are determined, which can be traced back to: ${\displaystyle {\vec {J}}^{2}|j,m\rangle =\hbar ^{2}j(j+1)|j,m\rangle }$ ${\displaystyle {J}_{z}|j,m\rangle =\hbar m|j,m\rangle }$ ${\displaystyle j}$ ${\displaystyle m}$ ${\displaystyle {J}_{\pm }={J}_{x}\pm \mathrm {i} {J}_{y}}$ ${\displaystyle [{J}_{a},{J}_{b}]=\mathrm {i} \hbar \varepsilon _{abc}{J}_{c}}$

${\displaystyle [{J}_{+},{J}_{-}]=2\hbar {J}_{z}\ ,\quad [{\vec {J}}^{2},{J}_{\pm }]=0\ ,\quad [{J}_{z},{J}_{\pm }]=\pm \hbar {J}_{\pm }}$

Now the effect of the ladder operators on the state is to be examined: ${\displaystyle |j,m\rangle }$

${\displaystyle {\vec {J}}^{2}{J}_{\pm }|j,m\rangle ={J}_{\pm }{\vec {J}}^{2}|j,m\rangle =\hbar ^{2}j(j+1){J}_{\pm }|j,m\rangle \quad \Rightarrow \quad {J}_{\pm }|j,m\rangle =c_{\pm }|j,m'\rangle }$

${\displaystyle {\begin{aligned}{J}_{z}{J}_{\pm }|j,m\rangle &=\left({J}_{\pm }{J}_{z}+[{J}_{z},{J}_{\pm }]\right)|j,m\rangle =\left(\hbar m{J}_{\pm }\pm \hbar {J}_{\pm }\right)|j,m\rangle \\&=\hbar (m\pm 1){J}_{\pm }|j,m\rangle \quad \Rightarrow \quad {J}_{\pm }|j,m\rangle =c_{\pm }|j,m\pm 1\rangle \end{aligned}}}$

When a ladder operator is used, it does not change, but it is increased or decreased by 1 (therefore the terms ascent and descent operator are justified). The next step is to determine the constant . To this end, products from ladder operators are first brought up and back: ${\displaystyle j}$ ${\displaystyle m}$ ${\displaystyle c_{\pm }}$ ${\displaystyle {\vec {J}}^{2}}$ ${\displaystyle {J}_{z}}$

${\displaystyle {\vec {J}}^{2}={J}_{x}^{2}+{J}_{y}^{2}+{J}_{z}^{2}=\left[{\tfrac {1}{2}}({J}_{+}+{J}_{-})\right]^{2}+\left[{\tfrac {1}{2\mathrm {i} }}({J}_{+}-{J}_{-})\right]^{2}+{J}_{z}^{2}={\tfrac {1}{2}}\left({J}_{+}{J}_{-}+{J}_{-}{J}_{+}\right)+{J}_{z}^{2}}$

The commutator performs on as well . Inserting delivers: ${\displaystyle [{J}_{+},{J}_{-}]=2\hbar {J}_{z}}$ ${\displaystyle {J}_{+}{J}_{-}={J}_{-}{J}_{+}+2\hbar {J}_{z}}$ ${\displaystyle {J}_{-}{J}_{+}={J}_{+}{J}_{-}-2\hbar {J}_{z}}$

${\displaystyle {J}_{+}{J}_{-}={\vec {J}}^{2}-{J}_{z}^{2}+\hbar {J}_{z}\ ,\quad {J}_{-}{J}_{+}={\vec {J}}^{2}-{J}_{z}^{2}-\hbar {J}_{z}}$

Since all eigenstates should be normalized, the length of the vector is and can be determined via the normal square; this makes use of the fact that the ascending and descending operators are adjoint to one another : ${\displaystyle c_{\pm }}$ ${\displaystyle {J}_{\pm }|j,m\rangle }$ ${\displaystyle {J}_{+}^{\dagger }={J}_{-}}$

${\displaystyle {\begin{aligned}|c_{\pm }|^{2}&=\|c_{\pm }|j,m\pm 1\rangle \|^{2}=\|{J}_{\pm }|j,m\rangle \|^{2}=\langle j,m|{J}_{\pm }^{\dagger }{J}_{\pm }|j,m\rangle =\langle j,m|{J}_{\mp }{J}_{\pm }|j,m\rangle \\&=\langle j,m|{\vec {J}}^{2}-{J}_{z}^{2}\mp \hbar {J}_{z}|j,m\rangle =\hbar ^{2}j(j+1)-\hbar ^{2}m^{2}\mp \hbar ^{2}m\end{aligned}}}$

${\displaystyle |c_{\pm }|^{2}=\hbar ^{2}(j\mp m)(j\pm m+1)\quad \Rightarrow \quad c_{\pm }=\hbar {\sqrt {(j\mp m)(j\pm m+1)}}}$

Since the norm of a vector is non-negative, it must apply. It follows: ${\displaystyle (j\mp m)(j\pm m+1)\geq 0}$

${\displaystyle -j\leq m\leq j}$

Applying the ascending operator to the highest state , becomes ; one applies the relegation operator to will . In both cases the ladder breaks off and the zero vector is obtained : ${\displaystyle m=j}$ ${\displaystyle c_{+}=0}$ ${\displaystyle m=-j}$ ${\displaystyle c_{-}=0}$

${\displaystyle {J}_{+}|j,j\rangle =0\quad ,\quad {J}_{-}|j,-j\rangle =0}$

By applying ( ) the ascent operator to the state with , one arrives at : ${\displaystyle n}$ ${\displaystyle n\in \mathbb {N} _{0}}$ ${\displaystyle m=-j}$ ${\displaystyle m=j}$

${\displaystyle -j+n=j\quad \Rightarrow \quad j={\frac {n}{2}}\quad {\text{mit}}\quad n\in \mathbb {N} _{0}}$

Therefore the possible quantum numbers do not have to be negative whole numbers or half numbers. ${\displaystyle j}$

Orbital angular momentum operator

The orbital angular momentum operator represents a special realization of an angular momentum . This is defined as follows: ${\displaystyle {\vec {L}}=({L}_{x},{L}_{y},{L}_{z})}$

${\displaystyle {\vec {L}}={\vec {r}}\times {\vec {p}}={\vec {e}}_{i}\varepsilon _{ijk}{r}_{j}{p}_{k}}$

Here is the position operator , the momentum operator and the -th unit vector . The quantum number is usually not designated with , but with . The following applies ("the orbital angular momentum is perpendicular to the position vector and the momentum vector"). It follows that the quantum numbers are whole numbers : ${\displaystyle {\vec {r}}}$ ${\displaystyle {\vec {p}}}$ ${\displaystyle {\vec {e}}_{i}}$ ${\displaystyle i}$ ${\displaystyle j}$ ${\displaystyle l}$ ${\displaystyle {\vec {L}}\cdot {\vec {r}}={\vec {L}}\cdot {\vec {p}}=0}$ ${\displaystyle l}$ ${\displaystyle l=0,1,2,\ldots \ .}$

The eigenvectors can be identified in the position representation with the spherical surface functions (see below). ${\displaystyle Y_{lm}(\varphi ,\vartheta )=\langle \varphi ,\vartheta |lm\rangle }$

Location representation of the orbital angular momentum in Cartesian coordinates

For the momentum operator and position operator, respectively, apply in the position representation . Inserting this into results in : ${\displaystyle {\vec {p}}=-\mathrm {i} \hbar {\vec {\nabla }}}$ ${\displaystyle {\vec {r}}={\vec {r}}}$ ${\displaystyle {\vec {L}}={\vec {r}}\times {\vec {p}}}$ ${\displaystyle {\vec {\nabla }}=\left({\partial \over \partial x},{\partial \over \partial y},{\partial \over \partial z}\right)}$

${\displaystyle {L}_{x}=-\mathrm {i} \hbar \left(y{\partial \over \partial z}-z{\partial \over \partial y}\right)}$

${\displaystyle {L}_{y}=-\mathrm {i} \hbar \left(z{\partial \over \partial x}-x{\partial \over \partial z}\right)}$

${\displaystyle {L}_{z}=-\mathrm {i} \hbar \left(x{\partial \over \partial y}-y{\partial \over \partial x}\right)}$

Location representation of the orbital angular momentum in spherical coordinates

With the Nabla operator or the gradient in spherical coordinates , after executing the cross products, one initially obtains

${\displaystyle {\vec {L}}=-\mathrm {i} \hbar \left({\vec {e}}_{\varphi }{\frac {\partial }{\partial \theta }}-{\frac {{\vec {e}}_{\theta }}{\sin \theta }}{\frac {\partial }{\partial \varphi }}\right)}$

The Cartesian components of can now be read from the Cartesian components of the unit vectors and : ${\displaystyle {\vec {L}}}$ ${\displaystyle {\vec {e}}_{\varphi }}$ ${\displaystyle {\vec {e}}_{\theta }\!\,}$

${\displaystyle {L}_{x}=\mathrm {i} \hbar \left(\sin \varphi {\frac {\partial }{\partial \theta }}+\cot \theta \cos \varphi {\frac {\partial }{\partial \varphi }}\right)}$

${\displaystyle {L}_{y}=\mathrm {i} \hbar \left(-\cos \varphi {\frac {\partial }{\partial \theta }}+\cot \theta \sin \varphi {\frac {\partial }{\partial \varphi }}\right)}$

${\displaystyle {L}_{z}=-\mathrm {i} \hbar {\frac {\partial }{\partial \varphi }}}$

The last line shows that the component of the angular momentum is the generator of a rotation (with an angle ) around the axis. ${\displaystyle z}$ ${\displaystyle \varphi }$ ${\displaystyle z}$

${\displaystyle {L}_{\pm }={L}_{x}\pm \mathrm {i} {L}_{y}=\hbar \exp(\pm \mathrm {i} \varphi )\left(\pm {\frac {\partial }{\partial \theta }}+\mathrm {i} \cot \theta {\frac {\partial }{\partial \varphi }}\right)}$

${\displaystyle {\vec {L}}^{2}=-\hbar ^{2}\left({\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\sin \theta {\frac {\partial }{\partial \theta }}+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right)=-\hbar ^{2}\Delta _{\theta ,\varphi }}$

${\displaystyle \Delta ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}\,{\frac {\partial }{\partial r}}\right)-{\frac {{\vec {L}}^{2}}{\hbar ^{2}r^{2}}}}$

In position notation, the operator corresponds to the angular component of the Laplace operator (except for the constant ). The eigenfunctions of the angle component and thus of and are the spherical surface functions : ${\displaystyle {L}^{2}}$ ${\displaystyle \Delta _{\theta ,\varphi }}$ ${\displaystyle -\hbar ^{2}}$ ${\displaystyle {\vec {L}}^{2}}$ ${\displaystyle {L}_{z}}$ ${\displaystyle Y_{l,m}(\theta ,\varphi )}$

${\displaystyle {\vec {L}}^{2}Y_{l,m}(\theta ,\varphi )=\hbar ^{2}l(l+1)Y_{l,m}(\theta ,\varphi )}$

${\displaystyle {L}_{z}Y_{l,m}(\theta ,\varphi )=\hbar mY_{l,m}(\theta ,\varphi )}$

The quantum numbers and are limited to integer values: ${\displaystyle l}$ ${\displaystyle m}$

${\displaystyle l=0,1,2,\dots ,\quad m=-l,\dots ,l}$

The spherical surface functions form a complete orthonormal system of square-integrable functions on the unit sphere :

${\displaystyle \int _{0}^{\pi }\mathrm {d} \theta \int _{0}^{2\pi }\mathrm {d} \varphi \,\sin \theta \,Y_{l,m}^{*}(\theta ,\varphi )\,Y_{k,n}(\theta ,\varphi )=\delta _{l,k}\delta _{m,n}}$

Generating a rotation

The operator rotate the position coordinates by the angle around the z-axis: ${\displaystyle {R}_{z}(\varphi )}$ ${\displaystyle \varphi }$

${\displaystyle {R}_{z}(\varphi )\,\,\psi (x,\,y,\,z)=\psi (x\cos \varphi -y\sin \varphi ,\,x\sin \varphi +y\cos \varphi ,\,z)}$

For infinitesimally small angles of rotation the angle functions to first order in order to be developed (see also: infinitesimal rotations ) and also the wave function: ${\displaystyle \delta \varphi }$ ${\displaystyle \delta \varphi }$ ${\displaystyle \delta \varphi =0}$

${\displaystyle {\begin{aligned}{R}_{z}(\delta \varphi )\,\,\psi (x,\,y,\,z)&=\psi (x-y\delta \varphi ,\,y+x\delta \varphi ,\,z)\\&=\psi (x,\,y,\,z)-y\delta \varphi {\frac {\partial \psi (x,\,y,\,z)}{\partial x}}+x\delta \varphi {\frac {\partial \psi (x,\,y,\,z)}{\partial y}}\\&=\left[1+\delta \varphi \left(x{\frac {\partial }{\partial y}}-y{\frac {\partial }{\partial x}}\right)\right]\psi (x,\,y,\,z)\\&=\left[1+\delta \varphi {\frac {\mathrm {i} }{\hbar }}{L}_{z}\right]\psi (x,\,y,\,z)\end{aligned}}}$

In the last step the definition of the z-component of the angular momentum operator was used. Since is Hermitian , the infinitesimal rotation operator is unitary . ${\displaystyle {L}_{z}}$ ${\displaystyle {R}_{z}(\varphi )}$

In spherical coordinates, an infinitesimal rotation around the z-axis is analogous to the above:

${\displaystyle {\begin{aligned}{R}_{z}(\delta \varphi )\psi (r,\,\vartheta ,\,\varphi )&=\psi (r,\,\vartheta ,\,\varphi +\delta \varphi )=\psi (r,\,\vartheta ,\,\varphi )+\delta \varphi {\frac {\partial \psi (r,\,\vartheta ,\,\varphi )}{\partial \varphi }}\\&=\left[1+\delta \varphi {\frac {\partial }{\partial \varphi }}\right]\psi (r,\,\vartheta ,\,\varphi )=\left[1+\delta \varphi {\frac {\mathrm {i} }{\hbar }}{L}_{z}\right]\psi (r,\,\vartheta ,\,\varphi )\end{aligned}}}$

In order to generate a finite rotation from such an infinitesimal rotation, consider the following limit:

${\displaystyle {R}_{z}(\varphi )=\lim _{N\to \infty }\left[R_{z}\left({\frac {\varphi }{N}}\right)\right]^{N}=\lim _{N\to \infty }\left[1+{\frac {\varphi }{N}}{\frac {i}{\hbar }}{L}_{z}\right]^{N}=\exp \left(\varphi {\frac {\mathrm {i} }{\hbar }}{L}_{z}\right)}$

Since the rotation operator is composed of unitary operators , it is itself unitary. ${\displaystyle {R}_{z}(\varphi )}$ ${\displaystyle {R}_{z}(\delta \varphi )={R}_{z}(\varphi /N)}$

A rotation around any axis (with ) around the angle can generally be written as: ${\displaystyle {\vec {e}}}$ ${\displaystyle {\vec {e}}\cdot {\vec {e}}=1}$ ${\displaystyle \varphi }$

${\displaystyle {R}_{\vec {e}}(\varphi )=\lim _{N\to \infty }\left[1+{\frac {\varphi }{N}}{\frac {\mathrm {i} }{\hbar }}{\vec {e}}\cdot {\vec {L}}\right]^{N}=\exp \left(\!\varphi {\frac {i}{\hbar }}{\vec {e}}\cdot {\vec {L}}\right)}$

Spin operator

See also : Spin operator and basis states for spin 1/2

The spinning is a further degree of freedom of a quantum mechanical particle and describes the angular momentum in its rest system ( angular momentum ). In the case of point-shaped particles there is no classic analogue for this (and therefore also no position representation). The spin operator commutes with all other degrees of freedom of the particle, e.g. B. Momentum operator and orbital angular momentum operator. Unlike the orbital angular momentum, it does not have to be perpendicular to the momentum operator. The quantum numbers and can be whole or half numbers. In the most common case they have the values: ${\displaystyle {\vec {S}}=({S}_{x},{S}_{y},{S}_{z})}$ ${\displaystyle s}$ ${\displaystyle m_{s}}$

${\displaystyle s={\tfrac {1}{2}}\ ,\quad m_{s}=\pm {\tfrac {1}{2}}}$

All quarks and leptons are spin 1/2 particles, as are many composite particles such as protons and neutrons . However, there are also particles with a different spin, e.g. B. the photon and other exchange bosons with spin 1, the baryonic delta with = 3/2 etc. ${\displaystyle s}$

With spin 1/2, the two eigenstates are often referred to as "spin up" and "spin down".

${\displaystyle \left|{\tfrac {1}{2}},{\tfrac {1}{2}}\right\rangle =\left|+\right\rangle =\left|\uparrow \right\rangle \ ,\quad \left|{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle =\left|-\right\rangle =\left|\downarrow \right\rangle }$

These states satisfy the eigenvalue equations

${\displaystyle {S}^{2}\left|\pm \right\rangle ={\tfrac {3\hbar ^{2}}{4}}\left|\pm \right\rangle }$

${\displaystyle {S}_{z}\left|\pm \right\rangle =\pm {\tfrac {\hbar }{2}}\left|\pm \right\rangle }$

The ladder operators have the following effect on the eigenstates:

${\displaystyle {S}_{+}\left|-\right\rangle =\hbar \left|+\right\rangle \ ,\quad {S}_{+}\left|+\right\rangle =0}$

${\displaystyle {S}_{-}\left|+\right\rangle =\hbar \left|-\right\rangle \ ,\quad {S}_{-}\left|-\right\rangle =0}$

The spin components can be expressed using the ladder operators: ${\displaystyle {S}_{x},{S}_{y}}$

${\displaystyle {S}_{x}={\tfrac {1}{2}}\left({S}_{+}+{S}_{-}\right)\ ,\quad {S}_{y}={\tfrac {1}{2i}}\left({S}_{+}-{S}_{-}\right)}$

The matrix representation of the operators is often used, whereby the following column vectors ( spinors ) are assigned to the eigenstates :

${\displaystyle \left|+\right\rangle ={\begin{pmatrix}1\\0\end{pmatrix}}\ ,\quad \left|-\right\rangle ={\begin{pmatrix}0\\1\end{pmatrix}}}$

${\displaystyle {S}^{2}={\frac {3\hbar ^{2}}{4}}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}\ ,\quad {S}_{z}={\frac {\hbar }{2}}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}}}$

${\displaystyle {S}_{+}=\hbar {\begin{pmatrix}0&1\\0&0\end{pmatrix}}\ ,\quad {S}_{-}=\hbar {\begin{pmatrix}0&0\\1&0\end{pmatrix}}}$

${\displaystyle {S}_{x}={\frac {\hbar }{2}}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}\ ,\quad {S}_{y}={\frac {\hbar }{2}}{\begin{pmatrix}0&-i\\i&0\end{pmatrix}}}$

Finally be about the relationship

${\displaystyle {S}_{i}={\tfrac {\hbar }{2}}{\sigma }_{i}}$

the spin components are linked to the Pauli matrices . ${\displaystyle {\sigma }_{i}}$

Angular momentum operator and angular momentum vector

Alignment and direction quantization

The eigenstates are called aligned to the z-axis . The vector from the three expected values is parallel to the z-axis. The amount of this vector is and therefore does not reach the length of the angular momentum vector given by , even with maximum alignment ( ) . Correspondingly, it applies to the squares of the operators for the x and y components that their expected values cannot become smaller than . This is why the quantum mechanical angular momentum differs from a vector in three-dimensional space that is accessible to visualization: It cannot be parallel to any axis in the sense that its component along this axis is exactly as large as its magnitude or length. Nevertheless, in physical texts the maximum possible alignment is often referred to in simplified form as "parallel position". ${\displaystyle \vert j,m\rangle }$ ${\displaystyle \langle {\vec {J}}\rangle =(\langle {J}_{x}\rangle ,\,\langle {J}_{y}\rangle ,\,\langle {J}_{z}\rangle )=(0,0,\hbar m)}$ ${\displaystyle \hbar |m|}$ ${\displaystyle m=j}$ ${\displaystyle {\sqrt {\langle {\vec {J}}^{2}\rangle }}=\hbar {\sqrt {\langle {J}_{x}^{2}\rangle +\langle {J}_{y}^{2}\rangle +\langle {J}_{z}^{2}\rangle }}={\sqrt {j(j+1)}}}$ ${\displaystyle \langle {J}_{x}^{2}\rangle =\langle {J}_{y}^{2}\rangle ={\frac {1}{2}}\hbar ^{2}(j(j+1)-m^{2})}$ ${\displaystyle {\frac {1}{2}}\hbar j}$

For a vector in three-dimensional space, the angular distance to the z-axis results from , where is the length of the vector. Transferring this to the quantum mechanical angular momentum leads to ${\displaystyle {\vec {r}}=(x,y,z)}$ ${\displaystyle \vartheta }$ ${\displaystyle \cos \vartheta =z/|{\vec {r}}|}$ ${\displaystyle |{\vec {r}}|={\sqrt {{\vec {r}}^{2}}}}$

${\displaystyle \cos \vartheta ={\frac {\langle {J}_{z}\rangle }{\sqrt {\langle {\vec {J}}^{2}\rangle }}}={\frac {m}{\sqrt {j(j+1)}}}}$.

The discrete eigenvalues ​​of the z-component can therefore be visualized in such a way that the angular momentum vector can only assume certain angles to the z-axis in these states. This is called directional quantization. The smallest possible angle is given by . For large values ​​of the angular momentum tends towards zero. For the smallest (not vanishing) quantum mechanical angular momentum , however , there is what contradicts the descriptive description as "parallel position". The component of the angular momentum perpendicular to the z-axis has a well-defined eigenvalue in every state , as can be seen from . Only their direction in the xy-plane is completely indeterminate, because the expected values ​​of both the x-component and the y-component of the angular momentum are in themselves zero. ${\displaystyle m}$ ${\displaystyle \cos \vartheta _{\mathrm {min} }={\frac {j}{\sqrt {j(j+1)}}}\equiv {\sqrt {1-{\frac {1}{j+1}}}}}$ ${\displaystyle \vartheta }$ ${\displaystyle j={\frac {1}{2}}}$ ${\displaystyle \vartheta _{\mathrm {min} }>45^{\circ }}$ ${\displaystyle \vert j,m\rangle }$ ${\displaystyle {\sqrt {j(j+1)-m^{2}}}}$ ${\displaystyle {J}_{x}^{2}+{J}_{y}^{2}={\vec {J}}^{2}-{J}_{z}^{2}}$

Clear behavior when rotating and mirroring

The angular momentum operator corresponds in some respects to the descriptive picture of the classical angular momentum. In particular, when the coordinate system is rotated, it behaves just like any other vector; H. its three components along the new coordinate axes are linear combinations of the three operators along the old axes according to the same formulas as (e.g.) for the classical angular momentum vector. This also applies (in any state of the system under consideration) to the three expected values , which together form the vector expected value of . Therefore, the length of the expected value of the angular momentum vector remains the same for rotations of the coordinate system (or the state). ${\displaystyle {\vec {J}}=({J}_{x},{J}_{y},{J}_{z})}$ ${\displaystyle ({J'_{x}},{J'_{y}},{J'_{z}})}$ ${\displaystyle ({J_{x}},{J_{y}},{J_{z}})}$ ${\displaystyle \langle {\vec {J}}\rangle =(\langle {J}_{x}\rangle ,\,\langle {J}_{y}\rangle ,\,\langle {J}_{z}\rangle )}$ ${\displaystyle {\vec {J}}}$ ${\displaystyle \vert \langle {\vec {J}}\rangle \vert ={\sqrt {\langle {J}_{x}\rangle ^{2}+\langle {J}_{y}\rangle ^{2}+\langle {J}_{z}\rangle ^{2}}}}$

When the coordinate system is mirrored, the angular momentum operator and its expected value also behave exactly like the mechanical angular momentum vector. They remain the same, like all other axial vectors (e.g. angular velocity , magnetic field , magnetic dipole moment ), in contrast to polar vectors (such as position vector , velocity vector , momentum vector ), which change their sign when mirrored . Axial vectors are also called pseudo vectors . ${\displaystyle {\vec {J}}}$ ${\displaystyle \langle {\vec {J}}\rangle }$

States in contrast to perception

The amount of the expected value vector remains the same for all rotations and reflections of the system, but there are states of the same quantum number for quantum numbers, for which the vector has a different length and which therefore cannot be converted into one another by rotation and reflections. For example, in a state is the expected value and its amount . This can result in different values depending on the value of , except in the cases and . The length for is zero. The length zero results for the expected value of the angular momentum vector also in the case of states such as , provided that and differ by more than and thus continues to apply to the expected values . In such states, the system shows what is known as “alignment” (in German, “alignment”) (although the German word is often used in general for the case that the system is viewed on the basis of its eigenstates in relation to a previously selected z-axis should). ${\displaystyle \langle {\vec {J}}\rangle }$ ${\displaystyle j{\mathord {\geq }}1}$ ${\displaystyle \vert j,m\rangle }$ ${\displaystyle \langle {\vec {J}}\rangle {\mathord {=}}(m\hbar ,\,0,\,0)}$ ${\displaystyle \vert \langle {\vec {J}}\rangle \vert =\vert m\hbar \vert }$ ${\displaystyle m}$ ${\displaystyle j{\mathord {=}}0}$ ${\displaystyle j{\mathord {=}}{\tfrac {1}{2}}}$ ${\displaystyle m{\mathord {=}}0}$ ${\displaystyle \vert \langle {\vec {J}}\rangle \vert =0}$ ${\displaystyle \left(\vert j,m\rangle +\vert j,\,-m\rangle \right)}$ ${\displaystyle m}$ ${\displaystyle -m}$ ${\displaystyle 1}$ ${\displaystyle \langle {J}_{x}\rangle {\mathord {=}}\langle {J}_{y}\rangle {\mathord {=}}0}$ ${\displaystyle \vert j,m\rangle }$

In the case (see section "Spin 1/2 and three-dimensional vector" in the article Spin ), the expected value of the angular momentum operator in every possible given state has the length and a direction in space can be specified according to which the quantum number can be assigned to this state is. ${\displaystyle j{\mathord {=}}{\tfrac {1}{2}}}$ ${\displaystyle {\tfrac {1}{2}}\hbar }$ ${\displaystyle m{\mathord {=}}{\mathord {+}}{\tfrac {1}{2}}}$

Addition of angular momentum

One assumes two angular momentum operators and , each of which has the quantum numbers and or and . Each of these angular momenta has its own inherent space defined by the eigenvectors to or be spanned. The angular impulses exchange with each other . ${\displaystyle {\vec {J}}_{1}}$ ${\displaystyle {\vec {J}}_{2}}$ ${\displaystyle j_{1}}$ ${\displaystyle m_{1}}$ ${\displaystyle j_{2}}$ ${\displaystyle m_{2}}$ ${\displaystyle \left|j_{1},m_{1}\right\rangle }$ ${\displaystyle {\vec {J}}_{1}^{2},{J}_{1z}}$ ${\displaystyle \left|j_{2},m_{2}\right\rangle }$ ${\displaystyle {\vec {J}}_{2}^{2},{J}_{2z}}$ ${\displaystyle [{\vec {J}}_{1},{\vec {J}}_{2}]=0}$

Now the individual angular momentum is coupled to form a total angular momentum:

${\displaystyle {\vec {J}}={\vec {J}}_{1}+{\vec {J}}_{2}}$

Thus applies automatically . The states of the overall system form the product space ( tensor product ) of the states of the individual systems. The products of the basic states of the individual systems form a basis: ${\displaystyle {J}_{z}={J}_{1z}+{J}_{2z}}$ ${\displaystyle \left|j_{i},m_{i}\right\rangle }$

${\displaystyle \left|j_{1},m_{1}\right\rangle \otimes \left|j_{2},m_{2}\right\rangle \equiv \left|j_{1},m_{1};j_{2},m_{2}\right\rangle }$

However, these are (mostly) not eigenvectors of the total angular momentum , so that it has no diagonal shape in this basis. Therefore one goes from the complete set of commuting operators with the eigen-states to the complete set of commuting operators with the eigen-states . In the new basis, the total angular momentum again has a simple diagonal form: ${\displaystyle {\vec {J}}^{2}}$ ${\displaystyle {\vec {J}}_{1}^{2},{J}_{1z},{\vec {J}}_{2}^{2},{J}_{2z}}$ ${\displaystyle \left|j_{1},m_{1};j_{2},m_{2}\right\rangle }$ ${\displaystyle {\vec {J}}^{2},{J}_{z},{\vec {J}}_{1}^{2},{\vec {J}}_{2}^{2}}$ ${\displaystyle \left|J,M,j_{1},j_{2}\right\rangle }$

${\displaystyle {\vec {J}}^{2}\left|J,M,j_{1},j_{2}\right\rangle =\hbar ^{2}J(J+1)\left|J,M,j_{1},j_{2}\right\rangle }$

${\displaystyle {J}_{z}\left|J,M,j_{1},j_{2}\right\rangle =\hbar M\left|J,M,j_{1},j_{2}\right\rangle }$

The quantum numbers for the total angular momentum and can have the following values: ${\displaystyle J}$ ${\displaystyle M}$

${\displaystyle J=|j_{1}-j_{2}|,\ |j_{1}-j_{2}|+1,\dots ,j_{1}+j_{2}}$

${\displaystyle M=m_{1}+m_{2}=-J,\dots ,J}$.

The transition from the product base to the own base takes place via the following development (taking advantage of the completeness of the product base): ${\displaystyle \left|j_{1},m_{1};j_{2},m_{2}\right\rangle }$ ${\displaystyle \left|J,M,j_{1},j_{2}\right\rangle }$

${\displaystyle \left|J,M,j_{1},j_{2}\right\rangle =\sum _{m_{1},m_{2}}\left|j_{1},m_{1};j_{2},m_{2}\right\rangle \langle \ j_{1},m_{1};j_{2},m_{2}|J,M,j_{1},j_{2}\rangle }$

Where are the Clebsch-Gordan coefficients . ${\displaystyle \langle \ j_{1},m_{1};j_{2},m_{2}|J,M,j_{1},j_{2}\rangle }$

Spin-orbit coupling

→ Main article: Spin-orbit coupling

A 1/2 spin is coupled with an orbital angular momentum.

${\displaystyle {\vec {J}}={\vec {L}}+{\vec {S}}}$

The spin quantum numbers are limited to and , the orbital angular momentum quantum numbers are and . Thus the total angular momentum quantum number can only assume the following values: ${\displaystyle \,s={\tfrac {1}{2}}}$ ${\displaystyle \,m_{s}=\pm {\tfrac {1}{2}}}$ ${\displaystyle l\in \mathbb {N} _{0}}$ ${\displaystyle m_{l}=-l,\dots ,l\!\,}$ ${\displaystyle J}$

• for : ${\displaystyle \,l>0}$ ${\displaystyle J=l\pm {\tfrac {1}{2}}}$
• for : . ${\displaystyle \,l=0}$ ${\displaystyle J={\tfrac {1}{2}}}$

Each state of the total angular momentum basis is composed of exactly two product basis states . At given can only be. ${\displaystyle \left|J,M,l,s\right\rangle }$ ${\displaystyle M=m_{l}+m_{s}=m_{l}\pm {\tfrac {1}{2}}}$ ${\displaystyle m_{l}=M\mp {\tfrac {1}{2}}}$

${\displaystyle \left|l+{\tfrac {1}{2}},M,l,{\tfrac {1}{2}}\right\rangle =\alpha _{+}\left|l,M-{\tfrac {1}{2}};\,{\tfrac {1}{2}},+{\tfrac {1}{2}}\right\rangle +\beta _{+}\left|l,M+{\tfrac {1}{2}};\,{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle }$   For   ${\displaystyle J=l+{\tfrac {1}{2}}}$

${\displaystyle \left|l-{\tfrac {1}{2}},M,l,{\tfrac {1}{2}}\right\rangle =\alpha _{-}\left|l,M-{\tfrac {1}{2}};\,{\tfrac {1}{2}},+{\tfrac {1}{2}}\right\rangle +\beta _{-}\left|l,M+{\tfrac {1}{2}};\,{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle }$   For   ${\displaystyle J=l-{\tfrac {1}{2}}}$

The coefficients are determined from the requirement that the states are orthonormal:

${\displaystyle \alpha _{\pm }=\pm {\frac {\sqrt {l+{\tfrac {1}{2}}\pm M}}{\sqrt {2l+1}}}}$   For   ${\displaystyle \beta _{\pm }={\frac {\sqrt {l+{\tfrac {1}{2}}\mp M}}{\sqrt {2l+1}}}}$

As an example , the orbital angular momentum should be coupled with a spin . In the following, write in abbreviated form and for the product base . ${\displaystyle l=1\!\,}$ ${\displaystyle \,s={\tfrac {1}{2}}}$ ${\displaystyle \left|J,M,l{=}1,s{=}{\tfrac {1}{2}}\right\rangle =\left|J,M\right\rangle }$ ${\displaystyle \left|l{=}1,m_{l};s{=}{\tfrac {1}{2}},m_{s}{=}{\pm {\tfrac {1}{2}}}\right\rangle =\left|m_{l};{\mathord {\pm }}\right\rangle }$

For there is a quartet : ${\displaystyle J={\tfrac {3}{2}}}$

${\displaystyle \left|{\tfrac {3}{2}},+{\tfrac {3}{2}}\right\rangle =\left|1;+\right\rangle }$

${\displaystyle \left|{\tfrac {3}{2}},+{\tfrac {1}{2}}\right\rangle ={\sqrt {\tfrac {2}{3}}}\,\left|0;+\right\rangle \,+\,{\sqrt {\tfrac {1}{3}}}\,\left|1;-\right\rangle }$

${\displaystyle \left|{\tfrac {3}{2}},-{\tfrac {1}{2}}\right\rangle ={\sqrt {\tfrac {1}{3}}}\,\left|-1;+\right\rangle \,+\,{\sqrt {\tfrac {2}{3}}}\,\left|0;-\right\rangle }$

${\displaystyle \left|{\tfrac {3}{2}},-{\tfrac {3}{2}}\right\rangle =\left|-1;-\right\rangle }$

For there is a doublet : ${\displaystyle J={\tfrac {1}{2}}}$

${\displaystyle \left|{\tfrac {1}{2}},+{\tfrac {1}{2}}\right\rangle =-{\sqrt {\tfrac {1}{3}}}\,\left|0;+\right\rangle \,+\,{\sqrt {\tfrac {2}{3}}}\,\left|1;-\right\rangle }$

${\displaystyle \left|{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle =-{\sqrt {\tfrac {2}{3}}}\,\left|-1;+\right\rangle \,+\,{\sqrt {\tfrac {1}{3}}}\,\left|0;-\right\rangle }$

Spin-spin coupling

In the following two 1/2 spins are coupled.

${\displaystyle {\vec {S}}={\vec {S}}_{1}+{\vec {S}}_{2}}$

The spin quantum numbers are limited to and . Thus, the total spin quantum numbers and can only take the following values: ${\displaystyle s_{1,2}={\tfrac {1}{2}}}$ ${\displaystyle m_{s_{1,2}}=\pm {\tfrac {1}{2}}}$ ${\displaystyle S}$ ${\displaystyle M_{S}}$

• ${\displaystyle S=0}$ then ${\displaystyle M_{S}=0}$
• ${\displaystyle S=1}$ then ${\displaystyle M_{S}=-1,0,1}$

In the following, write in abbreviated form and for the product base ${\displaystyle \left|S,M_{S},{\tfrac {1}{2}},{\tfrac {1}{2}}\right\rangle =\left|S,M_{S}\right\rangle }$ ${\displaystyle \left|{\tfrac {1}{2}},\pm {\tfrac {1}{2}};{\tfrac {1}{2}},\pm {\tfrac {1}{2}}\right\rangle =\left|\pm ;\pm \right\rangle }$

For there is a triplet : ${\displaystyle S=1}$

${\displaystyle \left|1,1\right\rangle =\left|+;+\right\rangle }$

${\displaystyle \left|1,0\right\rangle ={\frac {1}{\sqrt {2}}}{\Big (}\left|+;-\right\rangle +\left|-;+\right\rangle {\Big )}}$

${\displaystyle \left|1,-1\right\rangle =\left|-;-\right\rangle }$

For there is a singlet : ${\displaystyle S=0}$

${\displaystyle |0,0\rangle ={\frac {1}{\sqrt {2}}}{\Big (}|+;-\rangle -|-;+\rangle {\Big )}}$

literature

• Nolting: Basic course Theoretical Physics 5/2. Quantum Mechanics - Methods and Applications . Springer Verlag, ISBN 3540260358