Rotation_operator_(quantum_mechanics) Knowpia

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

Quantum mechanical rotations edit

With every physical rotation $R$ , we postulate a quantum mechanical rotation operator $D(R)$ which rotates quantum mechanical states.

|\alpha \rangle _{R}=D(R)|\alpha \rangle

In terms of the generators of rotation,

D(\mathbf {\hat {n}} ,\phi )=\exp \left(-i\phi {\frac {\mathbf {\hat {n}} \cdot \mathbf {J} }{\hbar }}\right),

where

\mathbf {\hat {n}}

is rotation axis,

\mathbf {J}

is angular momentum, and

\hbar

is the reduced Planck constant.

The translation operator edit

The rotation operator $\operatorname {R} (z,\theta )$ , with the first argument $z$ indicating the rotation axis and the second $\theta$ the rotation angle, can operate through the translation operator $\operatorname {T} (a)$ for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the state $|x\rangle$ according to Quantum Mechanics).

Translation of the particle at position $x$ to position $x+a$ : $\operatorname {T} (a)|x\rangle =|x+a\rangle$

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):

\operatorname {T} (0)=1

\operatorname {T} (a)\operatorname {T} (da)|x\rangle =\operatorname {T} (a)|x+da\rangle =|x+a+da\rangle =\operatorname {T} (a+da)|x\rangle \Rightarrow \operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a+da)

Taylor development gives:

\operatorname {T} (da)=\operatorname {T} (0)+{\frac {d\operatorname {T} (0)}{da}}da+\cdots =1-{\frac {i}{\hbar }}p_{x}da

with

p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}

From that follows:

\operatorname {T} (a+da)=\operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a)\left(1-{\frac {i}{\hbar }}p_{x}da\right)\Rightarrow {\frac {\operatorname {T} (a+da)-\operatorname {T} (a)}{da}}={\frac {d\operatorname {T} }{da}}=-{\frac {i}{\hbar }}p_{x}\operatorname {T} (a)

This is a differential equation with the solution

\operatorname {T} (a)=\exp \left(-{\frac {i}{\hbar }}p_{x}a\right).

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 $[H,\operatorname {T} (a)]=0.$ This result means that linear momentum for the system is conserved.

In relation to the orbital angular momentum edit

Classically we have for the angular momentum $\mathbf {L} =\mathbf {r} \times \mathbf {p} .$ This is the same in quantum mechanics considering $\mathbf {r}$ and $\mathbf {p}$ as operators. Classically, an infinitesimal rotation $dt$ of the vector $\mathbf {r} =(x,y,z)$ about the $z$ -axis to $\mathbf {r} '=(x',y',z)$ leaving $z$ unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

{\begin{aligned}x'&=r\cos(t+dt)=x-y\,dt+\cdots \\y'&=r\sin(t+dt)=y+x\,dt+\cdots \end{aligned}}

From that follows for states:

\operatorname {R} (z,dt)|r\rangle =\operatorname {R} (z,dt)|x,y,z\rangle =|x-y\,dt,y+x\,dt,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|x,y,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|r\rangle

And consequently:

\operatorname {R} (z,dt)=\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)

Using

T_{k}(a)=\exp \left(-{\frac {i}{\hbar }}p_{k}a\right)

from above with

k=x,y

and Taylor expansion we get:

\operatorname {R} (z,dt)=\exp \left[-{\frac {i}{\hbar }}\left(xp_{y}-yp_{x}\right)dt\right]=\exp \left(-{\frac {i}{\hbar }}L_{z}dt\right)=1-{\frac {i}{\hbar }}L_{z}dt+\cdots

with

L_{z}=xp_{y}-yp_{x}

the

z

-component of the angular momentum according to the classical cross product.

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

{\begin{aligned}&\operatorname {R} (z,t+dt)=\operatorname {R} (z,t)\operatorname {R} (z,dt)\\[1.1ex]\Rightarrow {}&{\frac {d\operatorname {R} }{dt}}={\frac {\operatorname {R} (z,t+dt)-\operatorname {R} (z,t)}{dt}}=\operatorname {R} (z,t){\frac {\operatorname {R} (z,dt)-1}{dt}}=-{\frac {i}{\hbar }}L_{z}\operatorname {R} (z,t)\\[1.1ex]\Rightarrow {}&\operatorname {R} (z,t)=\exp \left(-{\frac {i}{\hbar }}\,t\,L_{z}\right)\end{aligned}}

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

For the spin angular momentum about for example the $y$ -axis we just replace $L_{z}$ with ${\textstyle S_{y}={\frac {\hbar }{2}}\sigma _{y}}$ (where $\sigma _{y}$ is the Pauli Y matrix) and we get the spin rotation operator

\operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).

Effect on the spin operator and quantum states edit

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

A'=PAP^{-1}

where

P

is the basis transformation matrix. If the vectors

b

respectively

c

are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle

t

between them. The spin operator

S_{b}

in the first basis can then be transformed into the spin operator

S_{c}

of the other basis through the following transformation:

S_{c}=\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)

From standard quantum mechanics we have the known results ${\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }$ and ${\textstyle S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle }$ where $|b+\rangle$ and $|c+\rangle$ are the top spins in their corresponding bases. So we have:

{\frac {\hbar }{2}}|c+\rangle =S_{c}|c+\rangle =\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle \Rightarrow

S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle ={\frac {\hbar }{2}}\operatorname {D} ^{-1}(y,t)|c+\rangle

Comparison with ${\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }$ yields $|b+\rangle =D^{-1}(y,t)|c+\rangle$ .

This means that if the state $|c+\rangle$ is rotated about the $y$ -axis by an angle $t$ , it becomes the state $|b+\rangle$ , a result that can be generalized to arbitrary axes.

References edit

L.D. Landau and E.M. Lifshitz: Quantum Mechanics: Non-Relativistic Theory, Pergamon Press, 1985
P.A.M. Dirac: The Principles of Quantum Mechanics, Oxford University Press, 1958
R.P. Feynman, R.B. Leighton and M. Sands: The Feynman Lectures on Physics, Addison-Wesley, 1965

Summary

Quantum mechanical rotations edit

The translation operator edit

In relation to the orbital angular momentum edit

Effect on the spin operator and quantum states edit

See also edit

References edit