Rotation operator (quantum mechanics)

This article concerns the rotation operator, as it appears in quantum mechanics.

The translation operator

The rotation operator $,mbox\{R\}(z,\; t)$, with the first argument $,z$ indicating the rotation axis and the second $,t\; =\; heta$ the rotation angle, is based on the translation operator $,mbox\{T\}(a)$, which is acting on the state $|x\; angle$ in the following manner:

:$mbox\{T\}(a)|x\; angle\; =\; |x\; +\; a\; angle$

We have:

:$,mbox\{T\}(0)\; =\; 1$

:$,mbox\{T\}(a)\; mbox\{T\}(da)|x\; angle\; =\; mbox\{T\}(a)|x\; +\; da\; angle\; =\; |x\; +\; a\; +\; da\; angle\; =\; mbox\{T\}(a\; +\; da)|x\; angle\; Rightarrow$

:$,mbox\{T\}(a)\; mbox\{T\}(da)\; =\; mbox\{T\}(a\; +\; da)$

Taylor development gives:

:$,mbox\{T\}(da)\; =\; mbox\{T\}(0)\; +\; frac\{dmbox\{T\{da(0)\}\; da\; +\; ...\; =\; 1\; -\; frac\{i\}\{h\}\; p\_x\; da$

with

:$,p\_x\; =\; i\; h\; frac\{dmbox\{T\{da(0)\}$

From that follows:

:$,mbox\{T\}(a\; +\; da)\; =\; mbox\{T\}(a)\; mbox\{T\}(da)\; =\; mbox\{T\}(a)left(1\; -\; frac\{i\}\{h\}\; p\_x\; da\; ight)\; Rightarrow$

:$,\; [mbox\{T\}(a\; +\; da)\; -\; mbox\{T\}(a)]\; /da\; =\; frac\{dmbox\{T\{da\}\; =\; -\; frac\{i\}\{h\}\; p\_x\; mbox\{T\}(a)$

This is a differential equation with the solution $,mbox\{T\}(a)\; =\; mbox\{exp\}left(-\; frac\{i\}\{h\}\; p\_x\; a\; ight)$.

Additionally, suppose a Hamiltonian $,H$ is independent of the $,x$ position. Because the translation operator can be written in terms of $,p\_x$, and $,\; [p\_x,H]\; =0$, we know that $,\; [H,mbox\{T\}(a)]\; =0$. This result means that linear momentum for the system is conserved.

In relation to the orbital angular momentum

Classically we have $,l\; =\; r\; imes\; p$. This is the same in quantum mechanics considering $,r$ and $,p$ as operators. An infinitesimal rotation $,dt$ about the z-axis can be expressed by the following infinitesimal translations:

:$,x\text{'}\; =\; x\; -\; y\; dt$

:$,y\text{'}\; =\; y\; +\; x\; dt$

From that follows:

:$,mbox\{R\}(z,\; dt)|r\; angle$$=\; mbox\{R\}(z,\; dt)|x,\; y,\; z\; angle$$=\; |x\; -\; y\; dt,\; y\; +\; x\; dt,\; z\; angle$$=\; mbox\{T\}\_x(-y\; dt)\; mbox\{T\}\_y(x\; dt)|x,\; y,\; z\; angle$$=\; mbox\{T\}\_x(-y\; dt)\; mbox\{T\}\_y(x\; dt)|r\; angle$

And consequently:

:$,mbox\{R\}(z,\; dt)\; =\; mbox\{T\}\_x\; (-y\; dt)\; mbox\{T\}\_y(x\; dt)$

Using $,T\_k(a)\; =\; expleft(-\; frac\{i\}\{h\}\; p\_k\; a\; ight)$ with $,k\; =\; x,y$ and Taylor development we get:

:$,mbox\{R\}(z,\; dt)\; =\; expleft\; [-\; frac\{i\}\{h\}\; (x\; p\_y\; -\; y\; p\_x)\; dt\; ight]$$=\; expleft(-\; frac\{i\}\{h\}\; l\_z\; dt\; ight)\; =\; 1\; -\; i/h\; l\_z\; dt\; +\; ...$

To get a rotation for the angle $,t$, we construct the following differential equation using the condition $mbox\{R\}(z,\; 0)\; =\; 1$:

:$,mbox\{R\}(z,\; t\; +\; dt)\; =\; mbox\{R\}(z,\; t)\; mbox\{R\}(z,\; dt)\; Rightarrow$:$,\; [mbox\{R\}(z,\; t\; +\; dt)\; -\; mbox\{R\}(z,\; t)]\; /dt\; =\; dmbox\{R\}/dt$$,=\; mbox\{R\}(z,\; t)\; [mbox\{R\}(z,\; dt)\; -\; 1]\; /dt$$,=\; -\; frac\{i\}\{h\}\; l\_z\; mbox\{R\}(z,\; t)\; Rightarrow$:$,mbox\{R\}(z,\; t)\; =\; expleft(-\; frac\{i\}\{h\}\; t\; l\_z\; ight)$

Similar to the translation operator, if we are given a Hamiltonian $,H$ which rotationally symmetric about the z axis, $,\; [l\_z,H]\; =0$ implies $,\; [mbox\{R\}(z,t),H]\; =0$. This result means that angular momentum is conserved.

For the spin angular momentum about the y-axis we just replace $,l\_z$ with $,s\_y\; =\; frac\{h\}\{2\}\; sigma\_y$ and we get the spin rotation operator $,mbox\{D\}(y,\; t)\; =\; expleft(-\; i\; frac\{t\}\{2\}\; sigma\_y\; ight)$.

Effect upon the spin operator and upon states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix $,A$ can be represented in another base through the basis transformation

:$,A\text{'}\; =\; P\; A\; P^\{-1\}$

where $,P$ is the transformation matrix. If $,b$ and $,c$ are perpendicular to the y-axis and the angle $,t$ lies between them, the spin operator $,S\_b$ can be transformed into the spin operator $S\_c$ through the following transformation:

:$,S\_c\; =\; mbox\{D\}(y,\; t)\; S\_b\; mbox\{D\}^\{-1\}(y,\; t)$

From standard quantum mechanics we have the known results $,S\_b\; |b+\; angle\; =\; frac\{h\}\{2\}\; |b+\; angle$ and $,S\_c\; |c+\; angle\; =\; frac\{h\}\{2\}\; |c+\; angle$. So we have:

:$,frac\{h\}\{2\}\; |c+\; angle\; =\; S\_c\; |c+\; angle\; =\; mbox\{D\}(y,\; t)\; S\_b\; mbox\{D\}^\{-1\}(y,\; t)\; |c+\; angle\; Rightarrow$

:$,S\_b\; mbox\{D\}^\{-1\}(y,\; t)\; |c+\; angle\; =\; frac\{h\}\{2\}\; mbox\{D\}^\{-1\}(y,\; t)\; |c+\; angle$

Comparison with $,S\_b\; |b+\; angle\; =\; frac\{h\}\{2\}\; |b+\; angle$

yields $,|b+\; angle\; =\; D^\{-1\}(y,\; t)\; |c+\; angle$.

This can be generalized to arbitrary axes.