Confusion regarding taking the Hermitian adjoint of an outer product in spectral decomposition

Question

From my notes${}^{\zeta}$, I have that,

So in short, I don't understand why, when we take the Hermitian conjugate of an outer product, say, $\lvert A_i \rangle\langle A_i \rvert$ it seems like we should just end up with an inner product, say, $\langle A_i \lvert A_i \rangle$ since I am just applying the result of eqn $(3.30)$ above, which sends $\lvert A_i \rangle$ to $\langle A_i \rvert$, and vice versa, which is tantamount to switching the order of them.

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So, a more careful analysis is required...

I don't understand how eqn $(4.15)$ follows from eqn $(4.14)$, specifically, I know that for Hermitian operators $\hat{A}^{\dagger}=\hat{A}^*=\hat A$, since the eigenvalues of Hermitian operators are real. I understand this because the Hermitian adjoint, $\dagger$ operation means to transpose first, $\intercal$, then take the complex conjugate. But since there is only one operator, $\hat{A}$, we must have the case that $\hat{A}^{\intercal}=\hat{A}$.

In an attempt to understand this I will use 2 different indices, $i,j$, (as the fact that the same index was being used in my notes may be the source of my confusion). Starting with $(4.14)$, but only focussing on the outer product part, $$\lvert A_i \rangle\langle A_j \rvert$$ So if I can demonstrate that $\lvert A_i \rangle\langle A_j \rvert$ is invariant under $\dagger$, then I will have understood it.

I start by taking the Hermitian conjugate $$\Bigl(\lvert A_i \rangle\langle A_j \rvert\Bigl)^{\dagger}\stackrel{\intercal}{=}\Bigl(\lvert A_i \rangle\langle A_j \rvert\Bigl)^{\intercal}=\langle A_j \lvert A_i \rangle\stackrel{*}=\color{blue}{\Bigl(\langle A_j \lvert A_i \rangle\Bigl)^{*}=\langle A_i \lvert A_j \rangle}\ne \lvert A_i \rangle\langle A_j \rvert$$

In the blue part above I am making use of eqn $(3.27)$, which essentially states that, $\Bigl(\langle \beta \mid \gamma\rangle\Bigl)^*=\langle \gamma \mid \beta\rangle$ for Hilbert space vectors $\beta, \gamma$.

So in short, I am asking why, under the Hermitian adjoint operation, eqn $(4.14)$, $$\hat {A}=\sum_i A_i\lvert A_i\rangle\langle A_i \rvert$$ doesn't become $$\sum_i {A_i}^* \langle A_i \mid A_i \rangle?$$

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${}^{\zeta}$ Sources from ICL dept. of physics

Answer 1

Let's first look at what happens if we take the Hermitian conjugate of a ket: $$|A_i\rangle^\dagger = \langle A_i|.$$ If we have two operators $C$ and $D$, and want to take the Hermitian conjugate of their product, we have to reverse their order and take the Hermitian conjugates of $C$ and $D$: $$(CD)^\dagger = D^\dagger C^\dagger$$ Now to your question: If we have a ket-bra and want to take the Hermitian conjugate, we again have to reverse the order, and then take the Hermitian conjugate of the ket and bra individually: $$(|A_i\rangle \langle A_j|)^\dagger = (\langle A_j|)^\dagger (|A_i\rangle)^\dagger = |A_j\rangle\langle A_i|$$ Does this answer your question?

Answer 2

The projector $P_{A_i}=|A_i\rangle\langle A_i|$ is an operator, and the scalar product $\langle A_i|A_i\rangle$ is a complex number. If you do the adjoint of an operator you should get another operator, not a number. Thinking of them in their matrix form, transposing a matrix and then complex conjugating each entry gives another matrix, not a number. Sure, you can apply $|A_i\rangle^\dagger=\langle A_i|$ and $\langle A_i|^\dagger=|A_i\rangle$, but you have to be careful with the order, so that after doing the adjoint you still get an operator.

Another way of seeing that in general $(|\alpha\rangle\langle\beta|)^\dagger=|\beta\rangle\langle\alpha|$ is this: by definition, $\langle\phi|O^\dagger|\psi\rangle$ is the scalar product of $O|\phi\rangle$ and $|\psi\rangle$. Therefore it is also the scalar product of $|\psi\rangle$ and $O|\phi\rangle$ complex conjugated

$$\langle\phi|O^\dagger|\psi\rangle=\langle\psi|O|\phi\rangle^*.\tag{1}$$

So in particular, for an operator of the form $O=|\alpha\rangle\langle\beta|$ we have

\begin{align} \langle\psi|O|\phi\rangle^*&=\left(\langle\psi|\alpha\rangle\langle\beta|\phi\rangle\right)^*\\ &=\langle\psi|\alpha\rangle^*\langle\beta|\phi\rangle^*\\ &=\langle\alpha|\psi\rangle\langle\phi|\beta\rangle\\ &=\langle\phi\underbrace{|\beta\rangle\langle\alpha|}_{\displaystyle O^\dagger}\psi\rangle\tag{2},\\ \end{align} so by comparison with $(1)$, $(|\alpha\rangle\langle\beta|)^\dagger=O^\dagger=|\beta\rangle\langle\alpha|$.

Answer 3

Notations:

\begin{align} A_i &: \text{ is a real number}.\\ \big\vert A_i \rangle &: \text{ a column vector}.\\ \langle A_i \big\vert =\big\vert A_i \rangle^\dagger &: \text{ a row vector and complex adjoint}. \\ \big\vert A_i \rangle \langle A_i \big\vert &: \text{ a projection matrix}.\\ \langle A_i \big\vert A_i \rangle &= \text{ a real number}. \end{align}

And note the transporse of product of two matrix (or vector, a special matrix): $$ \left(\mathbf{A} \mathbf{B}\right)^T = \mathbf{B}^T \mathbf{A}^T $$ After transporse, the order of product is reversed.

Therefore, for the matrix $\mathbf{M}$:

$$ \mathbf{M} = \sum_i \lambda_i \big\vert U \rangle \langle V \big\vert $$

It is transport:

\begin{align} \mathbf{M}^\dagger &= \left\{\sum_i \lambda_i^* \big\vert U \rangle \langle V \big\vert \right\}^\dagger\\ &= \sum_i \lambda_i^* \left(\langle V \big\vert\right)^\dagger \left(\big\vert U \rangle \right)^\dagger\\ &= \sum_i \lambda_i^* \big\vert V \rangle \langle U \big\vert \end{align}