MHB Why is there a * in the 2nd integral for self-adjoint ODEs?

[ognik]

Hi, my textbook claims $ <u|\mathcal{L}v> =\int_{a}^{b}u^*\mathcal{L}v \,dx = \int_{a}^{b} u(p_0u''+p_1u'+p2u) \,dx$, u,v matrices or functions

My only query is why $u^*$, and where did the * get to in the 2nd integral? I am used to $ <f|g>=\int_{a}^{b}f(x)g(x) \,dx $ ...

 

[Ackbach]

The $^*$ is necessary in complex inner product spaces, such as you encounter in quantum mechanics. It's necessary to preserve some of the properties we expect inner products to have. As for it not appearing in the second integral, I suspect you're dealing with a real inner product space there, in which case the conjugate is unnecessary.

 

[ognik]

thanks ... and is it always the Bra part that is conjugated? And it's nothing to do with a conjugated Ket vector becoming a Bra?

 

[Ackbach]

thanks ... and is it always the Bra part that is conjugated? And it's nothing to do with a conjugated Ket vector becoming a Bra?

Here you have to be careful. Physicists are quite consistent here - the bra is conjugated and the ket is not. Mathematicians are inconsistent; some authors are linear in the first term, others in the second. I think the physics notation is superior because of how it suggests what's actually going on.

As for a conjugated ket becoming a bra, that's called taking the Hilbert adjoint. Technically, the conjugate transpose of a bra is a ket, and vice versa. This process takes a vector from one inner product space, and sends it into the dual space. This process is independent of which vector - the bra or the ket - you conjugate in an inner product.

 

[ognik]

Useful info thanks - I'm trying to become a physicist, so I'll stick with conjugating the Bra. I had seen that they were conjugate transposes of each other which was a tad confusing, so now I know they CAN, but will by default conjugate the Bra with inner products - hopefully I haven't misunderstood anything?

How does using the Bra suggest what's going on please? I thought it had something to do with needing one vector to be a col, the other a row - which of course doesn't work with other than 1 x n matrices...

 

[Ackbach]

Well, if you stick a bra $\langle y|$ together with a ket $|x\rangle$ you get a bra-ket, or bracket: $\langle y|x\rangle$, which is the inner product. Or if you have an expression like $|x\rangle\langle y|$, and you "multiply" it with a ket $|z\rangle$, you get $|x\rangle\langle y|z\rangle$. So I like the Dirac notation - I think it's intuitive, and helpful.

 

[ognik]

Just confirming, when you said it suggests what's going on, were you referring to the inner product? My query was more about physicists preferring the bra being conjugated instead of the ket, does conjugating the bra imply something deeper?

 

[Ackbach]

Just confirming, when you said it suggests what's going on, were you referring to the inner product?

Yep!

My query was more about physicists preferring the bra being conjugated instead of the ket, does conjugating the bra imply something deeper?

It does not imply anything mathematically deeper. It makes computations with Dirac notation a bit more straight-forward, that's all.

 

[ognik]

Gotcha, sorry if I sometimes appear pedantic, I have discovered too many small (but often significant) gaps and inaccuracies in my maths and am determined to increase my overall 'mathematical maturity' as well as do the course :-)