1. Sep 13, 2011

### Alexis21

Hello everybody,

why is the adjoint of a bra-ket like this:

$< \phi | \psi >^+ = < \psi | \phi >$

Is it a definition or can it be derived somehow?

Thanks :)

2. Sep 13, 2011

### vanhees71

The scalar product in a (pre-)Hilbert space is a sesquilinear form, i.e., by definition

$$\langle \psi|\phi \rangle = \langle \phi|\psi \rangle^*$$

and

$$\langle \psi | \alpha \phi_1 + \beta \phi_2 \rangle = \alpha \langle \psi | \phi_1 \rangle + \beta \langle \psi | \phi_2 \rangle.$$

3. Sep 14, 2011

### Alexis21

Thank you!

4. Sep 14, 2011

### Fredrik

Staff Emeritus
When $\langle\psi|\phi\rangle$ denotes the inner product (or semi-inner product) of $\psi$ and $\phi$, what vanhees71 said is the complete answer. But if it denotes $\langle\psi|$ acting on $|\phi\rangle$, some elaboration is required. $\langle\psi|$ is defined as a function that takes kets to complex numbers. To be more specific, it's defined as the function such that takes $|\phi\rangle$ to $\big(|\psi\rangle,|\phi\rangle\big)$. (Here I'm using the $(\cdot,\cdot)$ notation for the inner product of two kets, to make things more readable). Now we can prove it like this:
$$\langle\psi|\phi\rangle^* =\big(\langle\psi|\big(|\phi\rangle\big)\big)^* =\big(|\psi\rangle,|\phi\rangle\big)^* =\big(|\phi\rangle,|\psi\rangle\big) =\langle\phi|\big(|\psi\rangle\big) =\langle\phi|\psi\rangle$$ The equality in the middle is the same identity that vanhees71 mentioned. As he said, it's part of the definition of an inner product.

5. Sep 14, 2011

### dextercioby

The bra-ket is not a scalar product, but a short-hand for the action of a linear functional on a vector which yields a complex/real scalar.