4 let p be an arbitrary projector and i be the identity...

4. Let $P$ be an arbitrary projector and $I$ be the identity operator. a) Show that $2P - I$ is a unitary operator. b) Consider an n-qubit Hilbert space, and let $x$ be an n-bit bitstring. Describe the action of $2|x\rangle\langle x| - I$ on any computational basis state $|y\rangle$. c) Let $|\phi\rangle$ be the equal superposition of computational basis state $|\phi\rangle = \frac{1}{\sqrt{2^n}} \sum_{x \in \{0, 1\}^n} |x\rangle$. What is the result of applying $2|\phi\rangle\langle\phi| - I$ on any computational basis state $|y\rangle$.

Transcript

00:01 So in this question, we've got a projection operator x is the outer product of some state x with itself.

00:08 And x is a normalized ket, so we have this relationship here.

00:14 So first of all, x squared is outer product of the x's and then product with another outer product of the x's.

00:24 But this is just the inner product of the x's times the outer product of the x's, because we can pull that inner product out of the outer product.

00:33 That inner product is normalized, so it's equal to 1, so we get this outer product again.

00:40 So that's equal to x.

00:42 So we have x squared equals x.

00:46 Part b, x dagger is the dagger of this outer product of the two xes.

00:55 But when we take a dagger of an outer product, all we do is we swap the order of the bra and the ket, and then we turn the bra into a ket and the ket into a bra.

01:03 But that just leaves us with x outer product x again.

01:07 Because we've got the same thing in each of them.

01:10 So this is equal to x.

01:12 So we have x dagger equals x as well.

01:16 So here all we've done is we've swapped around these two things.

01:23 So now we've shown those two properties and they define a projector.

01:29 Or they are fundamental properties of any projection operator.

01:33 So now we have x and y are projectors.

01:38 And we need to show that x y must, xy can be a projector only if can be a projector only if the commutator of x and y is equal to zero.

01:59 So to show this, we need to show that x y squared equals xy.

02:04 This must be the case.

02:07 So let's expand this out.

02:09 We have xy xy is equal to xy now what we can do is we can commutate the x and the y in here so this is x squared y squared minus x commutator x y y y sorry plus commutator xy times y um so that's the right the left hand side but we know that x squared is x and y squared is y so we have x y plus x commutator xy y...

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