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From Nielsen & Chuang (10th edition), page 69:

Suppose $A$ is any linear operator on a Hilbert space, $V$. It turns out that there exists a unique linear operator $A^\dagger$ on $V$ such that for all vectors $|v\rangle$, $|w\rangle \in V$,

$$ (|v, A|w\rangle)=(A^\dagger|v\rangle, |w\rangle). \tag{2.32} $$

This linear operator is known as the adjoint or Hermitian conjugate of the operator $A$. From the definition it is easy to see that $(AB)^\dagger = B^\dagger A^\dagger$. By convention, if $|v\rangle$ is a vector, then we define $|v\rangle^\dagger \equiv \langle v|$. With this definition it is not difficult to see that $(A|v\rangle)^\dagger = \langle v|A^\dagger$.

Well, to me it is difficult to see that $$ (A|v\rangle)^\dagger = \langle v|A^\dagger \tag1\label1 $$

at least without invoking $$ A^\dagger=\left(A^* \right)^T \tag2\label2 $$ which I don't want to do because the book haven't introduced \eqref{2} at this point!

I realise that by using the definition $|v\rangle^\dagger \equiv \langle v|$ and right multiplying it by $A^\dagger$ I get:

$$ |v\rangle^\dagger A^\dagger = \langle v| A^\dagger \tag3 $$

This is pretty close to \eqref{1} and I only need to show that

$$ |v\rangle^\dagger A^\dagger = (A|v\rangle)^\dagger \tag4\label4 $$

My first instinct was to use $(AB)^\dagger = B^\dagger A^\dagger$ here. However, this does not feel quite right because $A$ and $B$ are both linear operators but in \eqref{4} I'm dealing with a linear operator and a vector. I tried getting around this by going to the matrix representation of linear operators and extending the vector $|v \rangle$ into a matrix such as:

$$ B = \begin{bmatrix} \vert & \vert & \dots & \vert \\ |v \rangle & 0 & \dots & 0 \\ \vert & \vert & \dots & \vert \\ \end{bmatrix} $$

Then I could invoke $(AB)^\dagger = B^\dagger A^\dagger$ but I'm not sure what to do with this because at this point in the book we don't know that $B^\dagger = \left(B^* \right)^T$. Therefore, we don't know that the first row of $B^\dagger$ will be $\langle v|$. Does anybody know how to proceed?

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The application of $(AB)^\dagger = B^\dagger A^\dagger$ directly is indeed not quite right.

At first note that $$ (A|v\rangle,|w\rangle) = (|w\rangle, A|v\rangle)^* = (A^\dagger|w\rangle, |v\rangle)^* = (|v\rangle , A^\dagger|w\rangle) $$ BTW, from this you can immediately deduce $(A^\dagger)^\dagger = A$.

Now for all $|w\rangle$ we have $$ (A|v\rangle)^\dagger |w\rangle = (A|v\rangle,|w\rangle) = (|v\rangle,A^\dagger|w\rangle) = $$ $$ = \langle v | \big(A^\dagger|w\rangle\big) = \langle v | A^\dagger|w\rangle = \big(\langle v | A^\dagger \big)|w\rangle $$ Since it's true for all $|w\rangle$ we can deduce the required $(A|v\rangle)^\dagger= \langle v | A^\dagger$.

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