# Is there a simplified formula for the adjoint of the outer product of ket and bra?

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?

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 $$\rho^\dagger = \big( |a\rangle \langle b| \big)^\dagger = |b\rangle\langle a|$$.

• \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? Feb 16 at 1:21

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 $$(|v\rangle\!\langle w|)^\dagger = |w\rangle\!\langle v|$$.

• Ah I see. Yes I should think like this. Feb 16 at 11:36