Squaring a Bra: Understanding the Order of <\phi|^2 and <\phi|\phi>

[ultimateguy]

Which of the following is true?

$$<\phi|^2 = <\phi|\phi>$$
$$<\phi|^2 = |\phi><\phi|$$

Or does the order even matter?

 

[genneth]

I'm not entirely sure you can "square a bra" like that. But the order certainly matters.

 

[ultimateguy]

Well I guess I should have said how do you multiply a bra by its complex conjugate, can you stick in the front or the back?

 

[genneth]

If you want a scalar, use option one. If you want an operator, use option two. Neither are called "squaring" -- and they're completely different things.

 

[michael879]

<phi|^2 = <phi|<phi| (pretty much meaningless)
<phi|phi> = 1 (or some other constant)
|phi><phi| = P_phi (an operator)

 

[OOO]

<phi|^2 = <phi|<phi| (pretty much meaningless)

I don't think the tensor product is meaningless. One can build multi-particle states from it.

 

[ultimateguy]

Ok well more specifically, my problem is turning this:

$$(E_n - E_{n'}) <\phi_{n'}|X|\phi_n> = \frac{i\hbar}{m} <\phi_{n'}|P|\phi_n>$$

into this:

$$\displaystyle\sum_{n'}^{} (E_n - E_{n'})^2 |<\phi_{n'}|X|\phi_n>|^2 = \frac{\hbar^2}{m^2} <\phi_n|P^2|\phi_n>$$

using the closure relation. So basically everything is getting squared, but I don't know how to handle the part with the P and it's respective bra and ket. And what is confusing me as well are the two $$\phi_n$$'s intead of a $$\phi_{n'}$$ in there.

 

[malawi_glenn]

is X and P the position and momentum operator?

What have you tried so far?

 

[Gokul43201]

Ok well more specifically, my problem is turning this:

$$(E_n - E_{n'}) <\phi_{n'}|X|\phi_n> = \frac{i\hbar}{m} <\phi_{n'}|P|\phi_n>$$

into this:

$$\displaystyle\sum_{n'}^{} (E_n - E_{n'})^2 |<\phi_{n'}|X|\phi_n>|^2 = \frac{\hbar^2}{m^2} <\phi_n|P^2|\phi_n>$$

using the closure relation. So basically everything is getting squared, but I don't know how to handle the part with the P and it's respective bra and ket. And what is confusing me as well are the two $$\phi_n$$'s intead of a $$\phi_{n'}$$ in there.

Keep the following in mind, and give it a shot:

$$\langle \phi _1 | A | \phi _2 \rangle ^* = \langle \phi _2 | A^{\dagger} | \phi _1 \rangle$$

If $\{ \phi _i \}$ is a complete basis, then $\sum _i |\phi _i \rangle \langle \phi _i |= \mathbf{1}$

You want to multiply every complex number in the upper expression by its complex conjugate and then sum over n'.

PS: Before doing this, you might want to review the math a little bit. It seems you are unfamiliar with some of the basics of the algebra of a linear vector space.

For instance:

Well I guess I should have said how do you multiply a bra by its complex conjugate, can you stick in the front or the back?

A bra is not a complex number (scalar) and hence, can not have a complex conjugate. It does, however, exhibit a dual correspondence to a ket living in a ket space that is dual to the bra space that your bra comes from.