In quantum mechanics, Kramers' theorem or Kramers' degeneracy theorem states that for every energy eigenstate of a time-reversal symmetric system with half-integer total spin, there is another eigenstate with the same energy related by time-reversal. In other words, the degeneracy of every energy level is an even number if it has half-integer spin. The theorem is named after Dutch physicist H. A. Kramers. In theoretical physics, the time reversal symmetry is the symmetry of physical laws under a time reversal transformation:
T : t ↦ − t . {\displaystyle T:t\mapsto -t.}
If the Hamiltonian operator commutes with the time-reversal operator, that is
[ H , T ] = 0 , {\displaystyle [H,T]=0,}
then, for every energy eigenstate | n ⟩ {\displaystyle |n\rangle } , the time reversed state T | n ⟩ {\displaystyle T|n\rangle } is also an eigenstate with the same energy. These two states are sometimes called a Kramers pair. In general, this time-reversed state may be identical to the original one, but that is not possible in a half-integer spin system: since time reversal reverses all angular momenta, reversing a half-integer spin cannot yield the same state (the magnetic quantum number is never zero).
Mathematical statement and proof In quantum mechanics, the time reversal operation is represented by an antiunitary operator T : H → H {\textstyle T:{\mathcal {H}}\to {\mathcal {H}}} acting on a Hilbert space H {\textstyle {\mathcal {H}}} . If it happens that T 2 = − 1 {\textstyle T^{2}=-1} , then we have the following simple theorem: If T : H → H {\textstyle T:{\mathcal {H}}\to {\mathcal {H}}} is an antiunitary operator acting on a Hilbert space H {\textstyle {\mathcal {H}}} satisfying T 2 = − 1 {\textstyle T^{2}=-1} and v {\textstyle v} a vector in H {\textstyle {\mathcal {H}}} , then T v {\textstyle Tv} is orthogonal to v {\textstyle v} .
Proof By the definition of an antiunitary operator, ⟨ T u , T w ⟩ = ⟨ w , u ⟩ {\textstyle \langle Tu,Tw\rangle =\langle w,u\rangle } , where u {\textstyle u} and w {\textstyle w} are vectors in H {\textstyle {\mathcal {H}}} . Replacing u = T v {\textstyle u=Tv} and w = v {\textstyle w=v} and using that T 2 = − 1 {\textstyle T^{2}=-1} , we get ⟨ T 2 v , T v ⟩ = − ⟨ v , T v ⟩ = ⟨ v , T v ⟩ {\textstyle \langle T^{2}v,Tv\rangle =-\langle v,Tv\rangle =\langle v,Tv\rangle } , which implies that ⟨ v , T v ⟩ = 0 {\textstyle \langle v,Tv\rangle =0} . Consequently, if a Hamiltonian H {\textstyle H} is time-reversal symmetric, i.e., it commutes with T , {\textstyle T,} then all its energy eigenspaces have even degeneracy, since applying T {\textstyle T} to an arbitrary energy eigenstate | n ⟩ {\textstyle |n\rangle } gives another energy eigenstate T | n ⟩ {\textstyle T|n\rangle } that is orthogonal to the first one. The orthogonality property is crucial, as it means that the two eigenstates | n ⟩ {\textstyle |n\rangle } and T | n ⟩ {\textstyle T|n\rangle } represent different physical states. If, on the contrary, they were the same physical state, then T | n ⟩ = e i α | n ⟩ {\displaystyle T|n\rangle =e^{i\alpha }|n\rangle } for an angle α ∈ R {\displaystyle \alpha \in \mathbb {R} } , which would imply
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