State of multiple q-bits qiven that one was already measured

State of multiple q-bits qiven that one was already measured

by Jérémy Valentin Barghorn -
Number of replies: 8

Hello,

I have some problems to understand how we are supposed to compute the global state of let's say 3 q-bits given that one of them was already measured and observed in one state.

So for example if I have the state  | \Psi> = \frac{1}{\sqrt{2}} (|00>_{AB} \otimes |1>_{C} + |11>_{AB} \otimes |0>_{C} )

And it is given that C measures his q-bit in the basis { |0>, |1> } and the result of the measurement is |0>

What are the computations we are doing to find the resulting state ? I would compute the following :  I_{AB} \otimes  |0><0|  | \Psi >  

But I'm not sure if we are allowed to do it like this. And the result would be  | \Psi> = \frac{1}{\sqrt{2}} (|00>_{AB} \otimes |1>_{C} ) with the 1/sqrt(2) factor in the front. Is it right ?

I'm asking the question with the 1/sqrt(2) factor in the front because in the 2023 exam the result of this question is : 


And the result after Charlie's measurement is :  without the 1/sqrt(2) in the front. Why is this the case ?

If I'm wrong in the computation can you please explain how we are supposed to do it the right way ?

Thank you in advance for the help !
In reply to Jérémy Valentin Barghorn

Re: State of multiple q-bits qiven that one was already measured

by Nicolas Macris -
What are the computations we are doing to find the resulting state ? I would compute the following : 𝐼𝐴𝐵⊗ |0><0| |Ψ>
YES CORRECT. YOU WILL GET THIS WAY THE UN-NORMALIZED STATE. TO GET THE "TRUE" NORMALIZED STATE YOU SHOULD DIVIDE BY THE NORM AND THE 1/sqrt 2 WILL GO AWAY.

But I'm not sure if we are allowed to do it like this. And the result would be |Ψ>=12√(|00>𝐴𝐵⊗|1>𝐶) with the 1/sqrt(2) factor in the front. Is it right ?

WELL YOU SHOULD GET (1/sqrt 2)|11>x|0>.
AS SAID THEN YOU SHOULD NORMALIZE (STATES ARE NORMALIZED) AND SO THE 1/sqrt 2 GETS DIVIDED OUT.

SAME FOR LAST QUESTION.
In reply to Nicolas Macris

Re: State of multiple q-bits qiven that one was already measured

by Yago Pérez Pérez -
"TO GET THE "TRUE" NORMALIZED STATE YOU SHOULD DIVIDE BY THE NORM AND THE 1/sqrt 2 WILL GO AWAY."
In this case, what are we taking the norm of? If we take norm of |+>, we get 1 so the 1/sqrt2 wont go away.
In reply to Yago Pérez Pérez

Re: State of multiple q-bits qiven that one was already measured

by Nicolas Macris -
If psi is a state and P one of the posdible projectors then the output quantum state is

P |psi> / (|| P|psi> ||)

The un-normalised vector is

P |psi>
The true physical state is the normalized total vector. The 1/sqrt 2 WILL GO AWAY
In reply to Nicolas Macris

Re: State of multiple q-bits qiven that one was already measured

by Quentin Marguet -
Hello, I think the computation is the following :

\ket{\Phi} = \frac{1}{\sqrt{2}} \left( \ket{0}_{A} \otimes \ket{0}_{B} \otimes\ket{1}_C +\ket{1}_{A}\otimes\ket{1}_{B}\otimes \ket{0}_C \right) \\
\ket{\Psi} =
\frac{\frac{1}{\sqrt{2}}\left(I_{A} \otimes I_B \otimes \ket{0}\bra{0} \right)\ket{\Phi}}{ \|\left(\frac{1}{\sqrt{2}}I_{A} \otimes I_B \otimes \ket{0} \bra{0}\right)\ket{\Phi}\|}=
\frac{\frac{1}{\sqrt{2}}I_{A} \ket{0}_A \otimes I_{B} \ket{0}_B }{ \|\frac{1}{\sqrt{2}}I_{A} \ket{0}_A \otimes I_{B} \ket{0}_B \|} =
\frac{ \frac{1}{\sqrt{2}}\left(\ket{0}_A \otimes \ket{0}_B \right) }{ \|\frac{1}{\sqrt{2}}\left(\ket{0}_A \otimes \ket{0}_B \right) \|} =2 \frac{ \frac{1}{\sqrt{2}}\left(\ket{0}_A \otimes \ket{0}_B \right) }{ \|\ket{0}_A \otimes \ket{0}_B \|} = \ket{0}_A \otimes \ket{0}_B

please use the following website to view the LaTeX  https://latexeditor.lagrida.com/, the expression requires the package "braket" which is not included by default in Tex


In reply to Nicolas Macris

Re: State of multiple q-bits qiven that one was already measured

by Yago Pérez Pérez -

I tried using the formula you provided for normalisation, and I still get the 1/sqrt2 :((

Where did I go wrong in my calculation? Could you show a detailed step by step? 


In reply to Yago Pérez Pérez

Re: State of multiple q-bits qiven that one was already measured

by Jérémy Valentin Barghorn -
Hello,
I'm not sure about your calculations but how I managed to finally solve the problem was to use this formula we have seen in the course to normalize a state :

And with this we can find that : 

In reply to Jérémy Valentin Barghorn

Re: State of multiple q-bits qiven that one was already measured

by Quentin Marguet -

Hello, 

indead I have made a big mistake thank you for the correction

In reply to Jérémy Valentin Barghorn

Re: State of multiple q-bits qiven that one was already measured

by Yago Pérez Pérez -
Thanks, I realized my mistake, been computing too many probabilities so I forgot completely about the square root part