7
$\begingroup$

Suppose we have a qutrit with the state vector $|\psi\rangle = a_0|0\rangle + a_1|1\rangle + a_2|2\rangle$, and we want to project its state onto the subspace having the basis $\{|0\rangle,|2\rangle\}$, I know the projection operator would be written like: $1|0\rangle \langle0| + 0|1\rangle\langle 1| + 1|2\rangle\langle2|$.

I'm having a few confusions here. Does $|0\rangle \langle 0|$ represent a tensor product between $[1 \ 0 \ 0]^{T}$ and $[1 \ 0 \ 0]$ ? Or is it just matrix multiplication? Also, I thought that we must always be able to write a projection operator in the form $|\phi\rangle \langle \phi|$ where $|\phi\rangle$ is a possible state of a qutrit. But how to represent $1|0\rangle \langle0| + 0|1\rangle\langle 1| + 1|2\rangle\langle2|$ in the form $|\phi\rangle \langle \phi|$?

$\endgroup$

2 Answers 2

8
$\begingroup$

Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication?

You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as the matrix multiplication (more precisely, Kronecker product) of the vectors representing them.

Also, I thought that we must always be able to write a projection operator in the form |ϕ⟩⟨ϕ|

Not necessarily. A projector will have that form if and only if it projects onto a one-dimensional space (that is, it projects onto a pure state). More general projections, like the one you mention, do not have this feature, and that is totally fine.

Indeed, also the identity matrix is a (trivial) projection, and it certainly cannot be written as $|\psi\rangle\langle\psi|$ for any pure state $|\psi\rangle$.

$\endgroup$
1
  • 1
    $\begingroup$ It's the same in this case, but I always thought of $|0\rangle\langle 0|$ as normal matrix product. $\endgroup$
    – Ali
    Jun 10, 2018 at 7:28
3
$\begingroup$

A projection operator $P$ has two key properties: $$ P^\dagger=P\qquad P^2=P $$ A particularly simple instance of a projection operator is a rank 1 projector, $P=|\phi\rangle\langle\phi|$, which you can easily see satisfies the two properties given that $|\phi\rangle$ is a normalised state, so $\langle\phi|\phi\rangle=1$.

To see what rank the projector is, simply evaluate $\text{rank}(P)=\text{Tr}(P)$. In your state example of $P=|0\rangle\langle 0|+|2\rangle\langle 2|$, you can see that the rank is 2, so it cannot be written as a rank 1 projector $|\phi\rangle\langle\phi|$.

$\endgroup$

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.