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I was reading about measurements and got to some operator like this: $$\left| 0\rangle \langle 0\right| $$

Is there any form I can apply when I have to calculate $$ \left( \left| 0\rangle \langle 0\right| \right) ^{+} $$

How can I simplify these expressions?

My work:

$$\left( \left| 0\rangle \langle 0\right| \right) ^{+}=\langle 0\left| ^{+}\right| 0\rangle ^{+}$$

Is this the way?

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Given that $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ then

$$ \rho = |0\rangle \langle 0| = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$

Thus, $\rho^\dagger = \big( |0\rangle \langle 0| \big)^\dagger = \rho $.

In general, given $\rho = |a\rangle\langle b| $ then $\rho^\dagger = \big( |a\rangle \langle b| \big)^\dagger = |b\rangle\langle a| $.

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    $\begingroup$ $$\begin{aligned}p^{+}=\left( \left| a\rangle \langle b\right| \right) ^{+}=\langle b\left| ^{+}\right| a\rangle ^{+}=\left| b\rangle \langle a\right| \\ .\end{aligned}$$ you mean this? Can you not jump that step? $\endgroup$
    – user27286
    Feb 16 at 1:21
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I think for these types of calculations it helps to use a more standard linear algebraic notation.

Given some finite-dimensional vector space $V$, let $v,w\in V$ be some vectors. Their outer product, in this context, is the linear operator denoted with $v w^\dagger$. It's worth noting that $w^\dagger$ is in this context also often denoted with $w^*$. This is the linear operator defined as $$(vw^\dagger)(x)=\langle w,x\rangle v, \qquad\forall x\in V,$$ where $\langle u,v\rangle\in\mathbb C$ denotes the inner product in the space.

In bra-ket notation, you write $v$ as $|v\rangle$ and $w^\dagger$ as $\langle w|$.

The adjoint (equivalently, the Hermitian conjugate) of $vw^\dagger\sim |v\rangle\!\langle w|$ can then be computed as simply the Hermitian conjugate of the corresponding matrix (more precisely, of the matrix representing the corresponding linear operator). The matrix elements of $vw^\dagger$ are $(vw^\dagger)_{ij}=v_i \bar w_j$, thus

$$(vw^\dagger)^\dagger_{ij} = \overline{(vw^\dagger)_{ji}} = \bar v_j w_i = (wv^\dagger)_{ij}.$$

This shows that $(vw^\dagger)^\dagger=(wv^\dagger)$, i.e. in bra-ket notation, that $(|v\rangle\!\langle w|)^\dagger = |w\rangle\!\langle v|$.

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  • $\begingroup$ Ah I see. Yes I should think like this. $\endgroup$
    – user27286
    Feb 16 at 11:36

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