I was reading these lecture notes (NB: PDF):
For spin-1/2, the rotation operator R(s)α(n)=exp(−iα2→σ⋅ˆn)
can be written as an explicit 2×2 matrix. This is accomplished by expanding the exponential in a Taylor series: exp(−iα2→σ⋅ˆn)=1−iα2(→σ⋅ˆn)+12!(iα2)2(→σ⋅ˆn)2−13!(iα2)3(→σ⋅ˆn)3+⋯Note that (→σ⋅ˆn)2=(→σ⋅ˆn)(→σ⋅ˆn)=ˆn⋅ˆn+iσ(ˆn׈n)=1Thus, the Taylor series becomes exp(−iα2→σ⋅ˆn)=1−iα2(→σ⋅ˆn)+12!(iα2)2(→σ⋅ˆn)2−13!(iα2)3(→σ⋅ˆn)3+⋯=[1−12!(α2)2+14!(α2)$+⋯]−i→σ⋅ˆn[(α2)−13!(α2)3+⋯]=cos(α2)−i→σ⋅ˆnsin(α2)
However, the part I don't understand is:
(→σ⋅ˆn)2=(→σ⋅ˆn)(→σ⋅ˆn)=ˆn⋅ˆn+iσ(ˆn׈n)=1
Why is that equal to 1? Where do the dot-product and cross-product come from? Note that the σ are Pauli spin matrices.
Answer
To show that (σ⋅n)2=n⋅n+iσ⋅(n×n)
consider writing the above as (σ⋅a)(σ⋅b)=∑jσjaj∑kσkbk=∑j∑k(12{σj,σk}+12[σj,σk])ajbk=∑j∑k(δjk+iϵjklσl)ajbk
where the 2nd line arises from using the anti-commutating and commutating relation for the matrices. In the third line, we have the Kronecker delta and Levi-Civita symbol. The result (1) follows from completing the math from (2) (that is, writing it in vector notation and replacing a and b with n).
The remainder is to show that this is equal to 1. For that, the following two hints should suffice:
- Note that for two vectors a and b, a×b=−b×a. What requirement is needed if b=a: a×a=?
- For the unit vector, e.g. n=(1,0)T, what is the dot product?
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