Nielsen and Chuang define Projectors as:

An operator $A$ whose adjoint is $A$ is known as a Hermitian or self-adjoint operator. An important class of Hermitian operators is the projectors. Suppose $W$ is a $k$-dimensional vector subspace of the $d$-dimensional vector space $V$. Using the Gram-Schmidt procedure it is possible to construct an orthonormal basis $\vert 1\rangle,\ldots ,\vert d\rangle$ for $V$ such that $\vert 1\rangle,\ldots ,\vert k\rangle$ is an orthonormal basis for $W$. By definition, $$P=\sum_{i=1}^k \vert i\rangle\langle i\vert\tag{2.35}$$

I have a basic doubt - (I think I am missing something simple, so please excuse my limited understanding): Since it is a sum over all orthonormal basis for $W$, should this not be the Identity, $\mathbb I$, according to the Completeness Relation?

  • 2
    $\begingroup$ Welcome to QCSE, please never use images when you can use text (use both if necessary). That way search engines can find this question easily, which could help other users. $\endgroup$
    – Mauricio
    Sep 6, 2021 at 15:28

1 Answer 1


$P$ is acting on the space $V$, projecting onto the subspace $W$. Yes, if it only acted on the subspace $W$, it would be identity, but it is acting on a larger space.

For example, think about a qubit, where $V$ is spanned by the basis states $\{|0\rangle,|1\rangle\}$. You can define a projector $P=|0\rangle\langle 0|$ which projects onto the space $W$ which, in this instance, is a single state $|0\rangle$.

  • $\begingroup$ Thanks Daft. If we just focus on the Vector Space W, what Im not able to understand is the difference between the sum above and the sum that adds to the Identity Matrix according to the Completeness Relation for W. $\endgroup$
    – sidharth
    Sep 6, 2021 at 9:41
  • $\begingroup$ @sidharth No, there isn't. The whole point is that this is identity (does nothing) on the space $W$ and is 0 on everything outside the space $W$. $\endgroup$
    – DaftWullie
    Sep 6, 2021 at 9:42

Your Answer

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

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