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

Hermitian Adjoint

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.

So, a more careful analysis is required...

Spectral decomposition

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 $$\sum_i {A_i}^* \langle A_i \mid A_i \rangle?$$

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

  • $\begingroup$ You should use MathJax to type the most important equations. It seems that you only need two or three equations, so please consider to type them instead of uploading (very large) images. $\endgroup$ Commented Jun 18, 2021 at 12:04
  • $\begingroup$ @Jakob "You should consider to use MathJax to write the most important equations", I already considered that and explicitly typed the ones I thought most important. If others are more important then please let me know and I will type those out also, thanks. $\endgroup$
    – BLAZE
    Commented Jun 18, 2021 at 12:06
  • $\begingroup$ You uploaded two large images with roughly ten equations, of which only two or three are important. $\endgroup$ Commented Jun 18, 2021 at 12:07
  • $\begingroup$ @Jakob What did I just write above? $\endgroup$
    – BLAZE
    Commented Jun 18, 2021 at 12:08

3 Answers 3


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?

  • $\begingroup$ Yes, this answers my question, sorry for the slow reply, been busy. The key word here is that Hermitian conjugation must be applied individually, what I was doing wrong in my question was applying it to the combined object. Many thanks. $\endgroup$
    – BLAZE
    Commented Jul 2, 2021 at 16:54
  • $\begingroup$ @NaturalLog Do you have to first reverse the order then take the Hermitian conjugate? $\endgroup$
    – Electra
    Commented Apr 4, 2023 at 10:28
  • 1
    $\begingroup$ @Electra There is no ordering to these operations. While you could think of the process first changing the order and then taking the Hermitian conjugate individually, a process where you first take the Hermitian conjugate individually and then reverse the order yields the same result. Apart from that, I find it misleading to assume these two steps to be separable. Just think of this process as happening in one go, without the chance to separate what happens throughout $\endgroup$
    – NaturalLog
    Commented Apr 5, 2023 at 11:43

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


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 $(1)$, $(|\alpha\rangle\langle\beta|)^\dagger=O^\dagger=|\beta\rangle\langle\alpha|$.



\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}

  • $\begingroup$ Thanks for your answer, does "transport"=transpose? $\endgroup$
    – BLAZE
    Commented Jun 18, 2021 at 12:46
  • $\begingroup$ Yes. The post corrected. $\endgroup$
    – ytlu
    Commented Jun 18, 2021 at 13:33

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