388 Quantum mechanics
which are (arbitrarily) referred to as “spin-up” and “spin-down,” respectively. The
spin-up and spin-down states are also sometimes denoted|α〉and|β〉, though we will
not make use of this nomenclature. Note that the operatorSˆ^2 =Sˆ·Sˆ=Sˆx^2 +Sˆ^2 y+Sˆz^2
is diagonal and, therefore, shares common eigenvectors withSˆz. These eigenvectors
are degenerate, however, having the eigenvalues(s+ 1) ̄h^2. Finally, if the Hamiltonian
is independent of the spin operator, then eigenvectors ofHˆare also eigenvectors ofSˆ^2
andSˆz, since all three can be simultaneously diagonalized.
How the physical states of identical particles are constructed depends on the spin
of the particles. Consider the example of two identical spin-sparticles. Suppose a
measurement is performed that can determine that one of the particles has anSˆz
eigenvalue ofma ̄hand the othermb ̄hsuch thatma 6 =mb. Is the state vector of the total
system just after this measurement|ma;mb〉≡|ma〉⊗|mb〉or|mb;ma〉≡|mb〉⊗|ma〉?
Note that, in the first state, particle 1 has anSˆzeigenvaluesma ̄h, and particle 2 has
mb ̄has theSˆzeigenvalue. In the second state, the labeling is reversed. The answer
is that neither state is correct. Since the particles are identical, the measurement is
not able to assign the particular spin states of each particle. In fact, the two states
|ma;mb〉and|mb;ma〉are not physically equivalent states. Two states|Ψ〉and|Ψ′〉
can only be physically equivalent if there is a complex scalarαsuch that
|Ψ〉=α|Ψ′〉, (9.4.4)and there is no such number relating|ma;mb〉to|mb;ma〉. Therefore, we need to con-
struct a new state vector|Ψ(ma,mb)〉such that|Ψ(mb,ma)〉is physically equivalent to
|Ψ(ma,mb)〉. Such a state is the only possibility for correctly representing the physical
state of the system immediately after the measurement. Let us take as an ansatz
|Ψ(ma,mb)〉=C|ma;mb〉+C′|mb;ma〉. (9.4.5)If we require that
|Ψ(ma,mb)〉=α|Ψ(mb,ma)〉, (9.4.6)
then
C|ma;mb〉+C′|mb;ma〉=α(C|mb;ma〉+C′|ma;mb〉), (9.4.7)
from which it can be seen that
C=αC′ C′=αC (9.4.8)or
C′=α^2 C′. (9.4.9)
The only solution to these equations isα=±1 andC=±C′. This gives us two possible
physical states of the system: a state that is symmetric (S) under an exchange ofSˆz
eigenvalues and one that is antisymmetric (A) under such an exchange. These states
are given by