Rotation operator (quantum mechanics)

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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)

\mathbf{\hat n} is rotation axis, and  \mathbf{J} is angular momentum.

The translation operator[edit]

The rotation operator \,\mbox{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 \,\mbox{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: \mbox{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):

\,\mbox{T}(0) = 1
\,\mbox{T}(a) \mbox{T}(da)|x\rangle = \mbox{T}(a)|x + da\rangle = |x + a + da\rangle = \mbox{T}(a + da)|x\rangle \Rightarrow
\,\mbox{T}(a) \mbox{T}(da) = \mbox{T}(a + da)

Taylor development gives:

\,\mbox{T}(da) = \mbox{T}(0) + \frac{d\mbox{T}(0)}{da} da + ... = 1 - \frac{i}{h}\ p_x\ da


\,p_x = i h \frac{d\mbox{T}(0)}{da}

From that follows:

\,\mbox{T}(a + da) = \mbox{T}(a) \mbox{T}(da) = \mbox{T}(a)\left(1 - \frac{i}{h} p_x da\right) \Rightarrow
\,[\mbox{T}(a + da) - \mbox{T}(a)]/da = \frac{d\mbox{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\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,\mbox{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 \,l = r \times p. This is the same in quantum mechanics considering \,r and \,p as operators. Classically, an infinitesimal rotation \,dt of the vector r=(x,y,z) about the z-axis to r'=(x',y',z) leaving z unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

\,x' = r \cos(t + dt) = x - y dt + ...
\,y' = r \sin(t + dt) = y + x dt + ...

From that follows for states:

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

And consequently:

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

Using \,T_k(a) = \exp\left(- \frac{i}{h}\ p_k\ a\right) from above with \,k = x,y and Taylor development we get:

\,\mbox{R}(z, dt) = \exp\left[- \frac{i}{h}\ (x p_y - y p_x) dt\right]= \exp\left(- \frac{i}{h}\ l_z dt\right) = 1 - \frac{i}{h} l_z dt + ...

with lz = x py - y px 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 \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 = d\mbox{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) = \exp\left(- \frac{i}{h}\ t\ l_z\right)

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) = \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' = P A P^{-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 = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t)

From standard quantum mechanics we have the known results \,S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle and \,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 = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t) |c+\rangle \Rightarrow
\,S_b \mbox{D}^{-1}(y, t) |c+\rangle = \frac{\hbar}{2} \mbox{D}^{-1}(y, t) |c+\rangle

Comparison with \,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. It is important, for instance, in Sakurai's Bell inequality.

See also[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

See also[edit]