# What is the adjoint of a ket-bra?

Let $T$ be a linear operator, then we can consider the rank-one operator

$$\vert Tx \rangle \langle y \vert.$$

I am wondering what is its adjoint operator, is it

$$\vert y \rangle \langle T^*x \vert?$$

• Replace $T^*$ for $T$ and your sipppsition is correct. – Valter Moretti Aug 15 '18 at 18:33
• – Cosmas Zachos Aug 15 '18 at 19:00

For any two vectors $|v\rangle$ and $|w\rangle$, the adjoint of $|w\rangle\langle v|$ is $|v\rangle\langle w|$.
So, the adjoint of $|Tx\rangle \langle y|$ is $|y\rangle \langle Tx|$.
• It might also be helpful to note that if $T|x\rangle = |Tx\rangle$ then $\langle Tx| = \langle x|T^†$ – Luke Pritchett Aug 15 '18 at 19:31