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' = x - y dt

:$,$y' = 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' = 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.