help needed in basic Qm

[imatfaal]

So the problem is as follows. you have two qubits in the state a|00> + b|11> - after measuring the first in the sign +- basis and after getting + you need to describe the state of the second qubit in standard 01 basis

 

So first off I changed the basis of the two qubits to the sign basis on the following lines (sorry for the plain test but \ket{} is not accepted by our latex renderer

 

|0> = 1/sqrt2 |+> + 1/sqrt2 |->

|1> = 1/sqrt2 |+> - 1/sqrt2 |->

 

a|00> = a(1/sqrt2 |+> + 1/sqrt2 |->)*(1/sqrt2 |+> + 1/sqrt2 |->)

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

 

b|11> = b(1/sqrt2 |+> - 1/sqrt2 |->)*(1/sqrt2 |+> 1 1/sqrt2 |->)

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

 

a|00> +b|11> = (a+b)/2 |++> + (a-b)/2 |-+> + (a-b)/2 |+-> + (a+b)/2 |-->

 

we know that first qubit was + and thus only interested in those terms and the descriptions of the second qubit of those interesting terms (***) is

 

(a+b)/2 |+> + (a-b)/2|->

 

convert back to standard 01 basis

(a+b)/2 (1/sqrt2 |0> +1/sqrt2 |1>) + (a-b)/2 (1/sqrt2 |0> - 1/sqrt2 |1>)

 

multiply out and you get

a/sqrt2 |0> + b/sqrt2 |1>

 

renormalise by dividing by sqrt of [(a/sqrt2)^2 + (b/sqrt2)^2 ] rearrange and cancel and you get

 

a/(a^2+b^2) |0> +b/(a^2+b^2)|1>

 

But I am told this is incorrect. I have tried changing the basis to a mixed one |+0> etc but I got the same answer.

 

I am pretty sure my mistake is where I have highlighted ***. I know I should rule out the terms which give - for first qubit but I am guessing the next step.

 

 

FYI this is a problem from an online course - so no answers please; I just want to know where I am screwing up and to get a prod in the right direction

[imatfaal]

I was wrong where my mistake was. Finally got it - I had not simplified out properly. We know from the initial question that as [latex]a|00\rangle+b|11\rangle [/latex] was already normalised that [latex]a^2+b^2 = 1[/latex] and thus you can simplify down to [latex]a|0\rangle+b|1\rangle [/latex]