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 p_x da with p_x = i h dT/da(0)

From that follows:

T(a + da) = T(a) T(da) = T(a)(1 - i/h p_x da) ⇒

[T(a + da) - T(a)]/da = dT/da = - i/h p_x T(a)

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

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〉 = T_x(-y dt) T_y(x dt)|x, y, z〉 = T_x(-y dt) T_y(x dt)|r〉

And consequently:

R(z, dt) = T_x(-y dt) T_y(x dt)

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

R(z, dt) = exp[- i/h (x p_y - y p_x) dt] = exp(- i/h l_z dt) = 1 - i/h l_z 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 l_z R(z, t) ⇒

R(z, t) = exp(- i/h t l_z)

For the spin angular momentum about the y-axis we just replace l_z with s_y = h/2 σ_y and we get the spin rotation operator D(y, t) = exp(- i t/2 σ_y).

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 S_b can be transformed into the spin operator S_c through the following transformation:

S_c = D(y, t) S_b D^-1(y, t)

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

h/2 |c+〉 = S_c |c+〉 = D(y, t) S_b D^-1(y, t) |c+〉 ⇒

S_b D^-1(y, t) |c+〉 = h/2 D^-1(y, t) |c+〉

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

This can be generalized to arbitrary axes.