# Homework Help: Operator Terminology

1. Mar 12, 2010

### Nezva

What does the asterisk mean?

2. Mar 12, 2010

### gabbagabbahey

The asterisk stand for complex conjugation.

3. Mar 13, 2010

### Nezva

In the case of $$\Psi=c$$1$$\Psi$$1$$+ c$$2$$\Psi$$2$$+ ... + c$$n$$\Psi$$n

And the operator A(hat) => A(hat)$$\Psi$$1 = a1$$\Psi$$1; A(hat)$$\Psi$$2 = a2$$\Psi$$2; A(hat)$$\Psi$$n = an$$\Psi$$n

Calculate: $$\left\langle\Psi\left|A(hat)\right|\right\Psi\rangle$$

4. Mar 13, 2010

### gabbagabbahey

Are you giving me a pop quiz?

You need to show us an attempt in order to receive help with your homework problem. Simply stating the problem does not qualify as an attempt.

5. Mar 13, 2010

### Nezva

I'm trying to substitute the psi function into the expectation value. I do not understand how to use the asterisk in this case. Especially if the constants in the function are complex... I'm trying to understand the concept and am unsure what to ask.

6. Mar 13, 2010

### Nezva

$$\int$$ c*n$$\Psi$$n$$\left| A(hat) \right| c$$n$$\Psi$$n

7. Mar 13, 2010

### gabbagabbahey

Okay, well when you say:

Do you mean $|\Psi\rangle=c_1|\psi_1\rangle+c_2|\psi_2\rangle+\ldots+c_n|\psi_n\rangle$ (abstract form), or do you mean $\Psi(\textbf{r})=c_1\psi_1(\textbf{r})+c_2\psi_2(\textbf{r})+\ldots+c_n\psi_n(\textbf{r})$ (all the wavefunctions are expanded in the position basis)?

More importantly, what do you know about $\{|\psi_1\rangle,|\psi_2\rangle,\ldots,|\psi_n\rangle\}$? For example, are they orthogonal? Normalized?

8. Mar 14, 2010

### Nezva

The $$\Psi$$is linear combination of n orthonormal eigenfunctions of the linear operator, A(hat).

What effect does a normalized function vs a 'orthonormal' function have on the 'expectation value'? Sorry I'm being thrown into this terminology very rudely. Any help is greatly appreciated.

9. Mar 14, 2010

### gabbagabbahey

Orthonormal means that the eigenfunctions are both orthogonal and normalized to unity.

In abstract form this means that $\langle\psi_i|\psi_j\rangle=\delta_{ij}$, where $$\delta_{ij}=\left\{\begin{array}{lr} 0, & i\neq j \\ 1, & i=j \end{array}\right.[/itex] is the Kronecker delta. When you expand the eigenfunctions in the position basis (i.e. $\psi(\textbf{r})=\langle\hat{\mathbf{r}}|\psi\rangle$), you get [tex]\langle\psi_i|\psi_j\rangle=\int_{-\infty}^{\infty}\psi_i^{*}(\textbf{r})\psi_j(\textbf{r})d\tau=\delta_{ij}$$

Start by calculating $\hat{A}|\Psi\rangle$...what do you get for that?