Project Ex1

Project Ex1

par Quentin Marguet,
Nombre de réponses : 6

Good evening, 

I was wondering in which way U was interacting with the first q-bit ? If we follow the wires going from left to right the operation acting on ket{0} are the two operations on the qbit just two Hadammard gates ?


Also what is the role of the q-bit 1 ? Which kind of control does it have on U since it is not present in the expression of the controlled unitary matrix ?

Thank you in advance

En réponse à Quentin Marguet

Re: Project Ex1

par Nicolas Macris,

Answer to first question: Yes. However after the controlled U the first qubit gets entangled with the rest and the application of the second Hadamard is non trivial. You have to write the math to see it clearly. If you did not have the control U you would have H^2|0>  = |0>, however because of the controlled op in the middle something different happens...

Answer to second question: well as said above you must write the math to see things clearly. After the first Hadamard the first qubit is left in a superposition of |0> + |1>. So the state of the circuit is 

|0>x |rest> + |1>x|rest>

and the control U will act non trivially on the |1>x|rest> part...

En réponse à Nicolas Macris

Re: Project Ex1

par Quentin Marguet,
Good afternoon,
first of all thank you for the answer.
The U matrix is therefore the "control matrix" and |0> the " target" q-bit or the other way around ?

Also I don't understand how we can do the math since U is a 2^n by 2^n matrix and |0> a 2 by 1 vector. Does that mean that we have to take the tensor product of both ? (but it would not make much sense since the measure only first the q-bit at the end)
Thank you in advance
En réponse à Quentin Marguet

Re: Project Ex1

par Nicolas Macris,

The contrary: the first top qubit is the control qubit and U acts on the target qubits. So U acts on the n target qubits and is indeed a 2^nx2^n matrix.

Best

N.M

En réponse à Nicolas Macris

Re: Project Ex1

par Quentin Marguet,
Good evening,

Thank you for the answer.

How does the U matrix get entangled with the |0> q-bit ? I really don't understand how the q-bits get entangled since it is a "control bit" and is not acted upon.

Also do we measure only the first q-bit or all the 1 + 2n q-bits on the {|0>, |1> } basis ?

How do we adapt the circuit for the mystery matrix use 2 (maybe) 3 q-bits as input and the ex 3 is based 5 q-bits as entry ?

In ex 4 should we generate angles between [0,2 Pi] or [0,Pi/2 ] as stated in the notebook ? Which one is the reference between the handout and the jupyter notebook ?

Thank you in advance
En réponse à Quentin Marguet

Re: Project Ex1

par Nicolas Macris,
Do the math! Uou will see that the first qbit gets entangled because before the control there is an Hadamard.

Only the first qu bit is measured in the computational basis.

For the mystery you have to adapt things: up to you to judge how many qubits you should use.

For the interval of angles you can take what you want basically. The notebook is a skeleton to guide you and you can slightly modify if you want.
En réponse à Nicolas Macris

Re: Project Ex1

par Quentin Marguet,
Good evening,

Thank you for the answer.

The thing that I did not understand was that the first q-bits and the "generalised" Bell State can be represented from the very beginning (even before the first Hadammard gate) using a tensor product,

Me and my partner were able to compute the overall state after the second Hadammard gate.
Now I am wondering how I can measure the first Q-bit since we have a entangled state ? How can we use the general born rule for that ?