Tensor Product Properties

Classical Mechanics Level 1

Two spin state are given by ∣ Ψ 1 ⟩ = ∣ ↑ ⟩ |\Psi_1\rangle = |\uparrow\rangle and ∣ Ψ 2 ⟩ = 1 2 ( ∣ ↑ ⟩ + ∣ ↓ ⟩ ) |\Psi_2 \rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle + |\downarrow\rangle) .

Which gives the correct tensor product state ∣ Ψ 1 ⟩ ⊗ ∣ Ψ 2 ⟩ |\Psi_1 \rangle \otimes |\Psi_2\rangle ?

Note : The notation ∣ ↑ ↓ ⟩ = ∣ ↑ ⟩ ⊗ ∣ ↓ ⟩ |\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle is a common shorthand for tensor products of spin states.

1 2 ∣ ↑ ↑ ⟩ \frac{1}{\sqrt{2}} \vert\uparrow\uparrow\rangle 1 2 ( ∣ ↑ ↑ ⟩ + ∣ ↓ ↑ ⟩ ) \frac{1}{\sqrt{2}} (\vert\uparrow\uparrow\rangle + \vert\downarrow\uparrow\rangle) 1 2 ∣ ↑ ↓ ⟩ \frac{1}{\sqrt{2}} \vert\uparrow\downarrow\rangle 1 2 ( ∣ ↑ ↑ ⟩ + ∣ ↑ ↓ ⟩ ) \frac{1}{\sqrt{2}} (\vert\uparrow\uparrow\rangle + \vert\uparrow\downarrow\rangle)

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
May 10, 2016

The distributivity of the tensor product gives the solution:

∣ Ψ 1 ⟩ ⊗ ∣ Ψ 2 ⟩ = ∣ ↑ ⟩ ⊗ 1 2 ( ∣ ↑ ⟩ + ∣ ↓ ⟩ ) = 1 2 ( ∣ ↑ ↑ ⟩ + ∣ ↑ ↓ ⟩ ) , |\Psi_1\rangle\otimes|\Psi_2\rangle = |\uparrow\rangle \otimes \frac{1}{\sqrt{2}} ( |\uparrow\rangle + |\downarrow\rangle) = \frac{1}{\sqrt{2}}( |\uparrow\uparrow\rangle + |\uparrow\downarrow\rangle),

as claimed.

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