# Rotation operator

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

## The translation operator

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

T(a)|x = |x + a

We have:

T(0) = 1
T(a) T(da)|x = T(a)|x + da = |x + a + da = T(a + da)|x
T(a) T(da) = T(a + da)

Taylor developement gives:

T(da) = T(0) + dT/da(0) da + ... = 1 - i/h px da with px = i h dT/da(0)

From that follows:

T(a + da) = T(a) T(da) = T(a)(1 - i/h px da) ⇒
[T(a + da) - T(a)]/da = dT/da = - i/h px T(a)

This is a differential equation with the solution T(a) = exp(- i/h px a).

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

## The rotation operator related to the orbital angular momentum

Classically we have l = r x p. This is the same in QM considering r and p as operators. An infinitesimal rotation dt about the z-axis can be expressed by the following infinitesimal translations:

x' = x - y dt
y' = y + x dt

From that follows:

R(z, dt)|r = R(z, dt)|x, y, z = |x - y dt, y + x dt, z = Tx(-y dt) Ty(x dt)|x, y, z = Tx(-y dt) Ty(x dt)|r

And consequently:

R(z, dt) = Tx(-y dt) Ty(x dt)

Using Tk(a) = exp(- i/h pk a) with k = x,y and Taylor developement we get:

R(z, dt) = exp[- i/h (x py - y px) dt] = exp(- i/h lz dt) = 1 - i/h lz dt + ...

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

R(z, t + dt) = R(z, t) R(z, dt) ⇒
[R(z, t + dt) - R(z, t)]/dt = dR/dt = R(z, t) [R(z, dt) - 1]/dt = - i/h lz R(z, t) ⇒
R(z, t) = exp(- i/h t lz)

Similar to the translation operator, if we are given a Hamiltonian H which rotationally symmetric about the z axis, [lz,H]=0 implies [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 lz with sy = h/2 σy and we get the spin rotation operator D(y, t) = exp(- i t/2 σy).

## The effect of the rotation operator on the spin operator and on states

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

A' = 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 Sb can be transformed into the spin operator Sc through the following transformation:

Sc = D(y, t) Sb D-1(y, t)

From standard QM we have the known results Sb |b+ = h/2 |b+ and Sc |c+ = h/2 |c+. So we have:

h/2 |c+ = Sc |c+ = D(y, t) Sb D-1(y, t) |c+
Sb D-1(y, t) |c+ = h/2 D-1(y, t) |c+

Comparison with Sb |b+ = h/2 |b+ yields |b+ = D-1(y, t) |c+.

This can be generalized to arbitrary axes.

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