Entangled or Not?

Classical Mechanics Level 1

Is the following state ∣ Ψ ⟩ |\Psi\rangle entangled?

∣ Ψ ⟩ = 1 2 ( ∣ ↓ ↓ ⟩ + ∣ ↓ ↑ ⟩ ) . |\Psi\rangle = \frac{1}{\sqrt{2}} (|\downarrow\downarrow\rangle + |\downarrow\uparrow\rangle).

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

Maybe Yes No

Matt DeCross by Matt DeCross , 27, USA

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2 solutions

Matt DeCross
Apr 26, 2016

The given state can be written as the tensor product of ∣ Ψ 1 ⟩ |\Psi_1\rangle and ∣ Ψ 2 ⟩ |\Psi_2\rangle with each of these two states given by:

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

Product states are by definition not entangled.

5 Helpful 0 Interesting 1 Brilliant 0 Confused
Tushar Kundu
Aug 30, 2020

As C11=0,C12=0,C21=1/√2,C22=1/√2 SO detM = 0, means not entangled

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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