0
$\begingroup$

enter image description here

The above picture is taken from a document. Can anyone please let me know the detailed internal calculations and how it is doing? like I don't understand how after the Tdagger gate when we apply H gate, the '01' state vanishes. So what are the step-by-step calculations is happening here? It will actually help me to underatnd other complex circuits also.Please help.

$\endgroup$

1 Answer 1

1
$\begingroup$

The state progression looks like this (in my own infra):

>>> qc.reg(2, 0)
<src.lib.state.Reg object at 0x7f14e2d98d00>
>>> qc.s(0)
>>> qc.h(1)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.71+0.00j prob: 0.50 Phase:   0.0
|01> (|1>):  ampl: +0.71+0.00j prob: 0.50 Phase:   0.0
>>> qc.h(0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|01> (|1>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|10> (|2>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|11> (|3>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
>>> qc.t(0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|01> (|1>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|10> (|2>):  ampl: +0.35+0.35j prob: 0.25 Phase:  45.0
|11> (|3>):  ampl: +0.35+0.35j prob: 0.25 Phase:  45.0
>>> qc.cx(1, 0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|01> (|1>):  ampl: +0.35+0.35j prob: 0.25 Phase:  45.0
|10> (|2>):  ampl: +0.35+0.35j prob: 0.25 Phase:  45.0
|11> (|3>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
>>> qc.tdag(0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|01> (|1>):  ampl: +0.35+0.35j prob: 0.25 Phase:  45.0
|10> (|2>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|11> (|3>):  ampl: +0.35-0.35j prob: 0.25 Phase: -45.0
>>> qc.h(0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.71+0.00j prob: 0.50 Phase:   0.0
|01> (|1>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|11> (|3>):  ampl: +0.00+0.50j prob: 0.25 Phase:  90.0
>>> qc.sdag(0)
>>> qc.psi.dump()
|00> (|0>):  ampl: +0.71+0.00j prob: 0.50 Phase:   0.0
|01> (|1>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0
|11> (|3>):  ampl: +0.50+0.00j prob: 0.25 Phase:   0.0

So the state after the $T^\dagger$ is: $$ |\psi\rangle = [ 0.5, 0.35+0.35i, 0.5, 0.35 - 0.35i ]^T $$

And indeed, multiplying the $H \otimes I$ now with $|\psi\rangle$ gives:

ops.Hadamard() * ops.Identity()
Operator([[ 0.70710677+0.j  0.        +0.j  0.70710677+0.j  0.        +0.j]
          [ 0.        +0.j  0.70710677+0.j  0.        +0.j  0.70710677+0.j]
          [ 0.70710677+0.j  0.        +0.j -0.70710677+0.j -0.        +0.j]
          [ 0.        +0.j  0.70710677+0.j -0.        +0.j -0.70710677+0.j]])

results in:

(ops.Hadamard() * ops.Identity())(s)
State([0.70710677+0.j         0.49497473+0.j         0.        +0.j
       0.        +0.49497473j])
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.