Elementary question about Dirac notation

[Ocifer]

Hello,

I'm in an introductory course about quantum computing. My math experience is fairly solid, but not very familiar with Dirac (bra-ket) notation. Just would like to clarify one thing:

In a single cubit space, we have |0 \rangle, and | 1 \rangle . I understand that these form an orthonormal basis for the the space, and can be represented by the standard basis vectors in matrix form.

My question is about multi-qubit systems, and about how to properly denote the conjugate transpose, or "bra" vector, related to the kets.

Suppose, I'm now in a two cubit space, which has the basis |0 \rangle |0 \rangle, |0 \rangle |1 \rangle, |1 \rangle |0 \rangle, \mathrm{and} |1 \rangle |1 \rangle, where two kets together implies a tensor product.

So suppose I have the vector |0 \rangle |1 \rangle = |01 \rangle. I get that this can be represented in matrix form by the column vector (0,1,0,0). From there its clear to me that if I take this vector's conjugate transpose, its "bra", it would simply get (0,1,0,0) as a row vector, and the scalar product of the two vectors would be 1.

My difficulty is moving from my intuition of matrix notation back to the Dirac notation.

When I take \mathrm{conjugate tranpose(}| 0 1 \rangle), would I get \langle 1 0 | ? That is to say, do I reverse the order of the 0s and 1s inside to take the bra vector?

I am leaning towards this being the case, because then we also get nice properties like

\langle 1 0 | 0 1 \rangle
= \langle 1 | \langle 0 | 0 \rangle |1 \rangle
= 1 \cdot \langle 1 | 1 \rangle
= 1 \cdot 1 = 1

instead of

\langle 01 | 01 \rangle = 1 (which is what is implied if we're not reversing the order of the 0s and 1s)

BOTTOM LINE QUESTION: Am I right in assuming that with Dirac notation, that when I take the conjugate transpose of some ket, \psi = |01001..1 \rangle, that I must reverse the order of the 0s and 1s in the bra vector? It seems to make sense, and it meshes with results about transposition of matrix products and tensor products that I already know from previous linear algebra work. I'm asking because I can't find it stated outright in my notes, and it's the only subtlety that seems to have tripped me up.

 

[Ben Niehoff]

No, the label inside the $\lvert \cdots \rangle$ is supposed to label the state, not act as some mnemonic for doing calculations. So

\big( \lvert 01 \rangle \big)^\dagger = \langle 01 \rvert
and

\langle abc \rvert = \big( \lvert abc \rangle \big)^\dagger = \big( \lvert a \rangle \lvert b \rangle \lvert c \rangle \big)^\dagger = \langle c \rvert \langle b \rvert \langle a \rvert
The reason for this is that for any normalized state whatsoever,

\langle abc \vert abc \rangle = 1

 

[Ocifer]

Thank you, sir.