Commuting
and noncommuting observables
Calculating commutators
Problem:
Consider the operators whose action is defined by the equations below:
O_{1}ψ(x) = x^{3}ψ(x)
O_{2}ψ(x) = x dψ(x)/dx
Find the commutator [O_{1}, O_{2}].
Solution:
 Concepts:
Commutator algebra
 Reasoning:
We are asked to find the commutator of two given operators.
 Details of the calculation:
[O_{1}, O_{2}]ψ(x) = O_{1}O_{2}ψ(x)  O_{2}O_{1}ψ(x)
= O_{1}xdψ(x)/dx  O_{2}x^{3}ψ(x) = x^{4}dψ(x)/dx
 xd(x^{3}ψ(x))/dx
= x^{4}dψ(x)/dx  3x^{3}ψ(x)  x^{4}dψ(x)/dx
= 3x^{3}ψ(x) = 3O_{1}ψ(x) for any ψ(x).
Therefore [O_{1}, O_{2}] = 3O_{1}.
Problem:
Let P and Q be two linear operators and let [P,Q] = iħ. Find
(a) [Q,P],
(b) [Q,P^{n}],
(c) [P,Q^{n}] .
Solution:
 Concepts:
Commutator algebra
 Reasoning:
We are asked to find several commutators.
The
product of two linear operators A
and B, written AB, is defined by
ABΨ> = A(BΨ>).
The order of the operators is important.
The commutator [A,B] is by
definition [A,B] = AB  BA.
Two useful identities using commutators are
[A,BC] = B[A,C] + [A,B]C and [AB,C]
= A[B,C] + [A,C]B.
Proof: [A,BC] = ABC  BCA + (BAC  BAC) = ABC + B[A,C]  BAC = B[A,C] +
[A,B]C.
 Details of the calculation:
(a) [Q,P] = iħ,
[Q,P^{2}] = P[Q,P] + [Q,P]P = 2iħP.
(b) [Q,P^{n+1}] = [Q,PP^{n}] = P[Q,P^{n}] + [Q,P]P^{n}.
If [Q,P^{n}] =iħnP^{n1},
then [Q,P^{n+1}] = PiħnP^{n1
}+ iħP^{n }= iħ(n+1)P^{n}.
[Q,P^{n}] = iħnP^{n1}
holds for n = 1 and n = 2. Therefore it holds for all n.
(c) [P,Q] = iħ, [P,Q^{2}]
= Q[P,Q] + [P,Q]Q = 2iħQ.
[P,Q^{n+1}] = [P,QQ^{n}] = Q[P,Q^{n}] + [P,Q]Q^{n}.
If [P,Q^{n}] = iħnQ^{n1},
then [P,Q^{n+1}] = QiħnQ^{n1
} iħQ^{n }= iħ(n+1)Q^{n}.
[P,Q^{n}] = iħnQ^{n1}
holds for n = 1 and n = 2. Therefore it holds for all n.
Problem:
Consider
a onedimensional system, with momentum operator p and position operator q.
(a) Show that
[q,p^{n}] = iħ n p^{n1}.
(b) Show that
[q,F(p)] = iħ ∂F/∂p, if
the function F(p) can be defined by a finite polynomial or convergent power series
in the operator p.
(c) Show that
[q,p^{2}F(q)] = 2iħ pF(q) if F(q)
is some function of q only.
Solution:
 Concepts:
Commutator algebra
 Reasoning:
We are asked to find several commutators.
 Details of the calculation:
(a) [q,p] = iħ,
[q,p^{2}] = p[q,p] + [q,p]p = 2iħp.
[q,p^{n+1}] = [q,pp^{n}] = p[q,p^{n}] + [q,p]p^{n}.
If [q,p^{n}] =iħnp^{n1},
then [q,p^{n+1}] = piħnp^{n1
}+ iħp^{n }= iħ(n+1)p^{n}.
[q,p^{n}] = iħnp^{n1}
holds for n = 1 and n = 2. Therefore it holds for all n.
(b) [q,F(p)] = [q,∑_{n}f_{n}p^{n}] = ∑_{n}f_{n}[q,p^{n}]
= ∑_{n}f_{n}niħp^{n1}] = iħ∂f(q)/∂q.
Similarly, [p,F(q)] = [p,∑_{n}f_{n}q^{n}] = ∑_{n}f_{n}[p,q^{n}]
= ∑_{n}f_{n}niħq^{n1}] = iħ∂f(q)/∂q.
(c) [q,p^{2}F(q)] = p^{2}[q,F(q)] + [q,p^{2}]F(q)]
= 0 + 2iħ pF(q).
FYI:
These formulas are used to proof the Ehrenfest's theorem.
(d/dt)<R> = <P>/m.
(d/dt)<P> = <∇U(R)>.
md^{2}<R>/dt^{2} = d<P>/dt = <∇U(R)>.
Problem:
The operators P and Q are represented by matrices
in some basis.
(a) Do P and Q commute?
(b) Find the normalized eigenvectors of P and Q.
Solution:
 Concepts:
Commuting operators
 Reasoning:
If the operators commute and the eigenvalues are not degenerate, they will
have the same eigenvectors.
 Details of the calculation:
(a)
The operators commute.
(b) To find the eigenvalues we calculate det(P  λI)
and set it equal to zero.
Eigenvalues of P: (1  λ)^{2}  4 = 0, λ_{1} = 3, λ_{2}
= 1.
Associated eigenvectors:
Eigenvalues of Q: (3  λ)^{2}  4 = 0, λ_{1} = 5, λ_{2}
= 1.
Associated eigenvectors:
P and Q have the same nondegenerate eigenvectors.
Problem:
The angular momentum operators {J_{x}, J_{y}, J_{z}} are central to quantum theory.
States are classified according to the eigenvalues of these operators when
J is conserved by the respective Hamiltonian H.
(a) What condition(s) is (are) necessary for all eigenstates of H to be
eigenstates of J?
An eigenstate of J is usually specified by j,m_{z}>,
where J^{2}j,m_{z}> = j(j + 1)ħ^{2}j,m_{z}>
and J_{z}j,m_{z}> = m_{z}ħj,m_{z}>.
(b) We can substitute J_{x} or J_{y} for J_{z}
in (a). However a state cannot be simultaneously an eigenstate of J_{z}
and J_{x}. Derive the commutation relation for the angular
momentum operators J_{x} and J_{z},
(i.e. [J_{x},J_{z}] = iħJ_{y})
from the definition of the linear momentum operator.
(c) Prove that it is indeed possible for a state to be simultaneously an eigenstate of
J^{2 }= J_{x}^{2 }+ J_{y}^{2 }+J_{z}^{2}
and J_{z}.
Solution:
 Concepts:
Commuting operators
 Reasoning:
Two commuting observables can be measured
simultaneously, i.e. the measurement of one does not cause loss of information
obtained in the measurement of the other. If we measure a complete set of commuting observables (C.S.C.O.), then the state
of the system after the measurement is one element of an unique eigenbasis. The
results of the measurement specify the state completely.
 Details of the calculation:
(a) It is possible to specify a common eigenbasis of two operators if they
commute. If one of the operators is non degenerate, then all of its
eigenvectors are also eigenvectors of the other operator. In order for all
eigenstates of H to be eigenstates of J^{2} and J_{z}
we need [J^{2},H] = 0 and [J_{z},H] = 0
and H is non degenerate. (States with different j or m_{z}
must have different energy eigenvalues E.)
(b) Assume the classical relationship J = r
× p.
J_{x} = yp_{z}  zp_{y}, J_{y} = zp_{x}
 xp_{z}, J_{z} = xp_{y}  yp_{x}.
For the quantum mechanical operators we then have
J_{x} = YP_{z}  ZP_{y}, J_{y} = ZP_{x}
 XP_{z}, J_{z} = XP_{y}  YP_{x}.
[J_{x},J_{z}] = [YP_{z}  ZP_{y},XP_{y}  YP_{x}]
= [YP_{z},XP_{y}] + [ZP_{y},YP_{x}]  [YP_{z},YP_{x}]
 [ZP_{y},XP_{y}]
0
0
= [YP_{z},X]P_{y} + X[YP_{z},P_{y}] + [ZP_{y},Y]P_{x}
+ Y[ZP_{y},P_{x}]
0
0
= Z[P_{y},Y]P_{x} + X[Y,P_{y}]P_{z} =
iħ(ZP_{x}
 XP_{z}) = iħJ_{y}.
(In general
[J_{i},J_{j}] = iħε_{ijk}J_{k}.)
(c)
J^{2} = J_{x}^{2} + J_{y}^{2} + J_{z}^{2}.
If [J^{2},J_{z}] = 0 then it is possible to
find simultaneous eigenstates.
[J^{2},J_{z}] = [J_{x}^{2},J_{z}] +
[J_{y}^{2},J_{z}] + [J_{z}^{2},J_{z}]
= J_{x}[J_{x},J_{z}] + [J_{x},J_{z}]J_{x}
+ J_{y}[J_{y},J_{z}] + [J_{y},J_{z}]J_{y
}= iħ(J_{x}J_{y} + J_{y}J_{x}) + iħ(J_{x}J_{y}
+ J_{y}J_{x}) = 0.
Problem:
Show that if any operator commutes with two of the components of an angular
momentum operator, it commutes with the third.
Solution:
 Concepts:
The commutation relations for the Cartesian components of any angular
momentum operator
 Reasoning:
Commutation relations are what defines a vector operator as a angular
momentum operator. We define angular momentum through
[J_{i},J_{j}] = ε_{ijk}iħJ_{k}.

Details of the calculation:
Let i ≠ j,k j ≠ k and let i, j, k be cyclic ( x, y, z or y, z, x or z, x, y ).
Then
[J_{i},J_{j}] = iħJ_{k}, J_{i}J_{j}
 J_{j}J_{i} = iħJ_{k}.
Let A be an operator.
Assume
[J_{i},A] = 0, [J_{j},A] = 0. Then [A, J_{k}]
= (1/iħ)[A,J_{i}J_{j}
 J_{j}J_{i}].
iħ[A,
J_{k}] = [A,J_{i}J_{j}]
 [A,J_{j}J_{i}]
= [A,J_{i}]J_{j}
+ J_{i}[A,J_{j}]
 [A,J_{j}]J_{i}
 J_{j}[A,J_{i}]
= 0.
Problem:
Consider the observables A = x^{2} and B = L_{z}.
Construct the generalized uncertainty relation
ΔA ΔB ≥ ½<i[A,B]>,
(i.e. find the commutator), for these observables.
Solution:
 Concepts:
Commutator algebra
 Reasoning:
L_{z} = (r×p)_{z} = xp_{y}  yp_{x},
[r_{i},p_{j}]
= iħδ_{ij}.
 Details of the calculation:
[x^{2},L_{z}] = x[x,L_{z}] + [x,L_{z}]x.
[x,L_{z}] = [x,xp_{y}]  [x,yp_{x}] = y[x,p_{x}]
= iħy.
[x^{2},L_{z}]= iħ[xy + yx] = 2iħxy.
Δx^{2}ΔL_{z} ≥ ħ<xy>.
Consequences
Problem:
Consider a system with Hamiltonian H.
Two observables A and B commute with H, [H,A] = [H,B] = 0, but do not commute
with each other, [A,B] ≠ 0. Show that the system has degenerate energy
level.
Solution:
 Concepts:
Commuting observables
 Reasoning:
If two observables commute, a common eigenbasis can be found. If they do not
commute, then no common eigenbasis exists.
 Details of the
calculation:
[H,A] = 0, an eigenstate with a nondegenerate eigenvalue of H is also an
eigenstate of A.
[H,B] = 0, an eigenstate with a nondegenerate eigenvalue of H is also an
eigenstate of B.
All nondegenerate eigenstates of H are also eigenstates of A and B.
[A,B] ≠ 0. There are eigenstates of H and A that are not eigenstates of B
and there are eigenstates of H and B that are not eigenstates of A. The eigenstates of H must be degenerate.
Problem:
Let A and B be two observables (Hermitian operators). In any state of the
system
ΔAΔB ≥ ½<i[A,B]>.
(a) Prove this generalized uncertainty principle.
[Hint: Let ψ> be any state vector and let A_{1 }= A  <A>I and B_{1
}= B  <B>I.
Let Φ> = A_{1}ψ> + ixB_{1}ψ> with x an arbitrary real
number. Use <ΦΦ> ≥ 0.]
Now consider a single particle in an eigenstate of L^{2} with wave
function Ψ(r,t).
(b) Calculate the commutators [sinφ, L_{z}] and [cosφ, L_{z}],
where φ is the azimuthal angle.
(c) Use these commutation relations and the result from part (a) to obtain
uncertainty relations between sinφ, L_{z} and cosφ, L_{z}.
Note: You can complete parts (b) and (c) without completing part (a).
Solution:

Concepts:
Generalized uncertainty principle, commutator algebra

Reasoning:
The hint leads to one way of proofing the generalized uncertainty principle.
There are certainly different ways to get the same result.

Details of the calculation:
(a) <ΦΦ> = <ψA_{1}^{2}ψ> + x^{2}<ψB_{1}^{2}ψ>
 ix<ψB_{1}A_{1}ψ> + ix<ψA_{1}B_{1}ψ>
= <A_{1}^{2}> + x^{2}<B_{1}^{2}> +
x<ψi[A_{1},B_{1}] ψ> = (ΔA)^{2 }+ x^{2}(ΔB)^{2
}+ x<i[A,B]>
for all x.
But <ΦΦ>≥ 0. Therefore (ΔA)^{2 }+ x^{2}(ΔB)^{2 }+
x<i[A,B]> ≥ = 0 for all x.
For this to be true we either need that the quadratic equation y = ax^{2
}+ bx + c either has no zeros, or is equal to zero for only one value of
x. (The equation describes a parabola. The vertex of the parabola can either
touch the xaxis or must lie above the xaxis.)
To find a root we set x = [b ± (b^{2 } 4ac)^{½}]/(2a)
.
If b^{2} < 4ac the equation has no roots. If b^{2 }= 4ac or
x = b/2a the equation is zero. We therefore need 4ac ≥ b^{2}.
Therefore ΔAΔB ≥ ½<i[A,B]>.
(b) In spherical coordinates the L_{z} operator is L_{z} =
(ħ/i)(∂/∂φ)
[sinφ, L_{z}]Ψ = (ħ/i) sinφ (∂Ψ/∂φ)  (ħ/i) cosφ Ψ  (ħ/i) sinφ
(∂Ψ/∂φ) =  (ħ/i) cosφ Ψ
for any wave function Ψ . Therefore [sinφ, L_{z}] =  (ħ/i) cosφ.
Similarly, [cosφ, L_{z}] = (ħ/i) sinφ.
(c) The generalized uncertainty relation: ΔAΔB ≥ ½<i[A,B]>.
Δsinφ ΔL_{z} ≥ ½ < i [ sinφ, L_{z}]> = (ħ/2) < cosφ >
Δcosφ ΔL_{z} ≥ ½ < i [ cosφ, L_{z}]> = (ħ/2) < sinφ >
If Ψ(r,t) is an eigenstate of L^{2}, i.e. if L is known, then Δsinφ
ΔL_{z} = Δcosφ ΔL_{z} ≥ 0.
Since L^{2} and L_{z} commute, the particle can be in an
eigenstate of L_{z}. Then ΔL_{z} = 0.
Since 1 ≥ cosφ ≥ 1 and 1 ≥ sinφ ≥ 1, (Δsinφ ΔL_{z})_{min} =
(Δcosφ ΔL_{z})_{min} = 0 for an eigenstate of L_{z}.