Bell States are maximally entangled states of two particles. Using two basis sta

Question

Bell States are maximally entangled states of two particles. Using two basis states |0> and |1>. one of the Bell states reads | psi > = (|0>1|0>2 + |1>11>2)/ = (|00> + |11>)/ . Express the Bell state | psi> using the states | plusminus >1,2 = (|0>)1,2 = |1>1,2)/ the bases states. Express the Bell State Psi> using the states | theta>1,2 = cos and as the bases states. Express the Bell State | Psi> using the states |R>1,2 = (|0)1,2 + i|1>1,2)/ and |L>1,2) = (|0>1,2 - i|1>1,2)/ as the basis states.

Explanation / Answer

|PSI> = (|00> + |11>)/2

(a) We have to find a linear combination that helps us achieve the Bell state above. For this case, we have:

|+>|+> = [(|0>1 + |1>1)/2][(|0>2 + |1>2)/2] = [|00> + |10> + |01> + |11>]/2

|->|-> = [(|0>1 - |1>1)/2][(|0>2 - |1>2)/2] = [|00> - |10> - |01> + |11>]/2

Let's add them together:

|+>|+> + |->|-> = [|00> + |10> + |01> + |11>]/2 + [|00> - |10> - |01> + |11>]/2

|+>|+> + |->|-> = (1/2)[2|00> + 2|11>] = |00> + |11>

So, we just need to divide that by 2

(1/2)(|+>|+> + |->|->)

(b) I will refer to -bar as '

|> = cos()|0> + sin()|1>

|'> = -sin()|0> + cos()|1>

|>|> = [cos()|0> + sin()|1>][cos()|0> + sin()|1>]

= cos2|00> + cos()sin()|01> + sin()cos()|10> + sin2|11>

|'>|'> = [-sin()|0> + cos()|1>][-sin()|0> + cos()|1>]

= sin2|00> - cos()sin()|01> - sin()cos()|10> + cos2|11>

Let's add them together and keep in mind that cos2 + sin2 = 1

|>|> + |'>|'> = cos2|00> + cos()sin()|01> + sin()cos()|10> + sin2|11> + sin2|00> - cos()sin()|01> - sin()cos()|10> + cos2|11>

= cos2|00> + sin2|00> + sin2|11> + cos2|11>

= |00> + |11>

Again, divide by 2

(1/2)(|>|> + |'>|'>)

(c) |R> = (|0> + i|1>)/2

|L> = (|0> - i|1>)/2

If we tried what we did earlier and do |R>|R> or |L>|L>, we'll have a negative coefficient on the |11> terms, which is unacceptable. So, let's try |R>|L> and |L>|R>

|R>|L> = [(|0> + i|1>)/2][(|0> - i|1>)/2]

= (|00> - i|01> + i|10> + |11>)/2

|L>|R> = [(|0> - i|1>)/2][(|0> + i|1>)/2]

= (|00> + i|01> - i|10> + |11>)/2

And let's add them up to eliminate the middle terms

|R>|L> + |L>|R> = (|00> - i|01> + i|10> + |11>)/2 + (|00> + i|01> - i|10> + |11>)/2

= [2|00> + 2|11>]/2

= |00> + |11>

And again, let's divide this by 2

(1/2)(|R>|L> + |L>|R>)

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