# Inner Product in this step of the working

1. Jan 21, 2014

### unscientific

Hi guys, I'm not sure how to evaluate this inner product at step (3.8)

I know that:

$\hat {H} |\phi> = E |\phi>$

$$<E_n|\frac{\hat H}{\hbar \omega} + \frac{1}{2}|E_n>$$

$$<E_n| \frac{\hat H}{\hbar \omega}|E_n> + <E_n|\frac{1}{2}|E_n>$$

I also know that $<\psi|\hat Q | \psi>$ gives the average value of observable to $\hat Q$. In this case, it's not $\psi$ but $E_n$, does the same principle hold?

Last edited: Jan 21, 2014
2. Jan 21, 2014

### George Jones

Staff Emeritus
Use equation (3.4).

3. Jan 21, 2014

### dextercioby

Hi George, I can only conclude that you figured out where he picked his 2 questions from. .

4. Jan 22, 2014

### unscientific

Okay, I got step 3.8. But For Step 3.9, I'm having a slight issue here:

$$E_n = <E_n|\hat H|E_n> = \frac{1}{2m}<E_n|(m\omega \hat x)^2 + \hat {p}^2|E_n>$$
$$= \frac{1}{2m}(m^2\omega^2)<E_n|\hat x \hat x|E_n> + \frac{1}{2m}<E_n|\hat p \hat p|E_n>$$

Using $\hat x|\phi> = x|\phi>$ and $\hat p |\phi> = p|\phi>$,

$$= \frac{1}{2m}(m^2\omega^2)<E_n|\hat x x|E_n> + \frac{1}{2m}<E_n|\hat p p|E_n>$$

Removing the last "hats" from $\hat p$ and $\hat x$ and Using orthogonality $<E_n|E_n> = 1$:

$$= \frac{1}{2m}\left ( (m\omega x)^2 + p^2 \right )$$

Is it wrong to assume that the energy eigenkets have norm = 1? That would be strange because later we show that $E_n = (n + \frac{1}{2})\hbar \omega$

And, why is there an additional factor of $\frac{1}{\omega}$ in the expression?

5. Jan 22, 2014

### dextercioby

Energy eigenkets for the harmonic oscillator do have norm =1.

6. Jan 22, 2014

### unscientific

Ok, then I have no idea what's gone wrong with my working..

7. Jan 22, 2014

### dextercioby

The energy eigenket is not an eigenket of either x nor p. You need the ladder operators to evaluate the matrix elements.

8. Jan 22, 2014

### unscientific

How did they get 3.9 then?

9. Jan 22, 2014

### dextercioby

<p E_n, p E_n> = ||p |E_n> ||^2 = <E_n, p^2 E_n>, because the eigenkets of energy are in the domain of both p and p^2, on which the 2 operators are essentially self-adjoint. p^2 is a positive operator, hence the inequality at the end.

The same goes for x.

10. Jan 22, 2014

### unscientific

I just don't get how $<E_n|\hat x \hat x|E_n> = |x|E_n>|^2$ and $<E_n|\hat p \hat p|E_n> = |p|E_n>|^2$

11. Jan 22, 2014

### ChrisVer

hmm because in the same way you have in the complex numbers that:
$z^{*} z = |z|^{2}$
In fact the ket can be interpreted as a vector on Hilber space, while the bra as its dual.
So in the case of this, you can write:

$< E_{n}| \hat{x} \hat{x} |E_{n}>= (\hat{x} |E_{n}>)^{t} \hat{x} |E_{n}>$
using that x operator is self adjoint. The same goes for p... with the "t" I denoted the adjoint conjugate operation

12. Jan 22, 2014

### unscientific

That is true, but $\hat x$ and $\hat p$ are operators.. so the x and p in the norm should have hats?

$<E_n|\hat x \hat x|E_n> = |\hat x|E_n>|^2$ and $<E_n|\hat p \hat p|E_n> = |\hat p|E_n>|^2$

13. Jan 22, 2014

### ChrisVer

they do have hats... the person who wrote the things in the image you posted, doesn't use hats so much...

You can get a feeling there must be hats because otherwise there would be no reason to use the eigenvectors in kets... (he'd get 1)

14. Jan 22, 2014

### unscientific

This clears things up a little, thanks!