# Normalization and expectation

1. Mar 14, 2015

### Maylis

Hello,

I am very confused how this is true? Where does this come from??
$$<f| \hat{Q}f> = (\sum_{n}a_{n}^{*} |\psi_{n}>)(\sum_{m}a_{m} \hat{Q} | \psi_{m}>)$$

thanks

2. Mar 14, 2015

### stevendaryl

Staff Emeritus
I think your formula is slightly wrong. If you have a complete set of orthonormal basis states $|\psi_m\rangle$, then you can write an arbitrary state $|f\rangle$ as a linear combination of basis states:

$|f\rangle = \sum_m a_m |\psi_m\rangle$

That being the case, you take the adjoint of both sides:

$|f\rangle^\dagger = \langle f | = \sum_m a_m^* \langle \psi_m|$

So as to not mix up the indices, let's relabel $m$ by $n$ in this expression:

$|f\rangle^\dagger = \langle f | = \sum_n a_n^* \langle \psi_n|$

Then it follows automatically that

$\langle f|\hat{O}|f\rangle = (\sum_n a_n^* \langle \psi_n|)\ \hat{O}\ ( \sum_m a_m |\psi_m\rangle)$

which can also be written as:
$\langle f|\hat{O}|f\rangle = (\sum_n a_n |\psi\rangle)^\dagger\ \hat{O}\ ( \sum_m a_m |\psi_m\rangle)$

Maybe you don't know what $\langle f|$ means? Well, you know that for ordinary wave functions, you define $\langle f|g\rangle$ to be the integral:

$\langle f|g\rangle = \int dx f^*(x) g(x)$

Then you can define $\langle f|$ as that operator that acts on $|g\rangle$ to produce $\langle f|g\rangle$

3. Mar 14, 2015

### Maylis

Ok, I was definitely unaware that $|f \rangle^{ \dagger} = \langle f|$. No wonder I was so confused over the definition of a hermition conjugate, how somehow moving the operator to the bra somehow made it a complex conjugate.You are right, I am very shaky about $\langle f|$. I understand the ket is a vector, but what is the bra? I know you say its an operator that acts on the vector g to produce the inner product, but that leaves me feeling icky inside. As in no intuition at all.

4. Mar 14, 2015

### stevendaryl

Staff Emeritus
Well, the simplest case is when there are only a finite number of states. For example, an electron's state, if you ignore the spatial part, is a two-state system: It can be spin-up or spin-down. So if we use matrices to represent the states, then we can have:

$|\psi_1\rangle = \left( \begin{array}\\ 1 \\ 0 \end{array} \right)$ which represents spin-up

$|\psi_2\rangle = \left( \begin{array}\\ 0 \\ 1 \end{array} \right)$ which represents spin-down

An arbitrary state $|f\rangle$ is a combination:

$|f\rangle = a_1 |\psi_1\rangle + a_2 |\psi_2\rangle = \left( \begin{array}\\ a_1 \\ a_2 \end{array} \right)$

Then $|f\rangle^\dagger$ is just $\left( \begin{array}\\ a_1^* & a_2^* \end{array} \right)$. You compute $|g\rangle^\dagger |f\rangle$ by matrix multiplication: If $|g\rangle$ is $\left( \begin{array}\\ b_1 \\ b_2 \end{array} \right)$, then

$|g\rangle^\dagger |f\rangle = \left( \begin{array}\\ b_1^* & b_2^* \end{array} \right)\ \left( \begin{array}\\ a_1 \\ a_2 \end{array} \right) = b_1^* a_1 + b_2^* a_2$

To generalize to an infinite, but countable, number of states,
$|g\rangle^\dagger |f\rangle = \sum_i b_i^* a_i$

To generalize to a continuous number of states,

$|g\rangle^\dagger |f\rangle = \int dx\ g(x)^* f(x)$

5. Mar 15, 2015

### Maylis

Thanks, that clears things up a lot.