Where Does the Equation for Normalization and Expectation Come From?

[gfd43tg]

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

 

[stevendaryl]

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

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

 

[gfd43tg]

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.

 

[stevendaryl]

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.

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^* &amp; 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^* &amp; 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)

 

[gfd43tg]

Thanks, that clears things up a lot.