If you have three things in a bra ket what does it mean?

[rwooduk]

say you have <ψ|x|ψ⟩ or <0|F|k⟩ where F is an operator, what does this actually mean? I understand C|ψ⟩ would be the operator C acting on PSI and <ψ1|ψ2⟩ is the inner product of two wavefunctions but what would a third term inbetween them mean?

thanks for any help

 

[stevendaryl]

Well, when an operator F acts on a state |\psi\rangle, it produces a new state: |\psi&#039;\rangle = F |\psi\rangle. So the expression \langle \phi | F | \psi \rangle is just the inner product of |\phi\rangle and F |\psi\rangle.

The nice thing about Hermitian operators is that the inner product of |\phi\rangle and F |\psi\rangle is the same as the inner product of F |\phi\rangle and |\psi\rangle. In \langle \phi | F | \psi \rangle, you can view F as acting to the left on \phi or to the right on \psi.

 

[rwooduk]

ahh okay many thanks! its the bra notation for the wavefunction in front of the operator that confuses me, could you explain how it differs from the wavefunction in the ket?

also if F can act left or right, is that like saying the two wavefunctions are overlapping (or bound) so it doesn't matter which the operator acts on first?

thanks again for the reply!

 

[Khashishi]

It doesn't matter which wavevector (the bra or the ket) the operator acts on, but that doesn't have anything to do with the wavefunctions overlapping or bound.

 

[dextercioby]

The true mathematical justification of the use of bras and kets cannot be formulated in the absence of rigged Hilbert spaces. But this state of the art, which many people wouldn't understand. To be 'sandwich'ed between a ket and a bra an operator needs to be selfadjoint. One then has that

$$\langle A\psi, \phi\rangle = \langle \psi, A\phi\rangle \equiv \langle \psi |A|\phi\rangle $$

 

[Avodyne]

An operator need not be self-adjoint to be sandwiched, e.g.,
$\langle m|a^\dagger|n\rangle = \sqrt{m}\delta_{m,n+1}$

 

[Haborix]

An operator need not be self-adjoint to be sandwiched, e.g.,
$\langle m|a^\dagger|n\rangle = \sqrt{m}\delta_{m,n+1}$

I think the opposition to sandwiching operators which aren't self-adjoint is that it is ambiguous notation. You have to think about acting to the left with the adjoint of the operator that you see in the middle. The notation generally employed by mathematicians avoids such ambiguities.

 

[bhobba]

The true mathematical justification of the use of bras and kets cannot be formulated in the absence of rigged Hilbert spaces

Very true.

But everyone into QM should have a smattering of knowledge of it:
http://physics.lamar.edu/rafa/cinvestav/second.pdf

Thanks
Bill

 

[Avodyne]

Nothing ambiguous about it. $\langle m|a^\dagger|n\rangle = \langle m|(a^\dagger|n\rangle) = (\langle m|a^\dagger)|n\rangle$, and $\langle m|a^\dagger =(a|m\rangle)^\dagger$.

 

[bhobba]

Nothing ambiguous about it. $\langle m|a^\dagger|n\rangle = \langle m|(a^\dagger|n\rangle) = (\langle m|a^\dagger)|n\rangle$, and $\langle m|a^\dagger =(a|m\rangle)^\dagger$.

The issue is in Rigged Hilbert spaces - care is required in exactly what one can apply an operator to.

Thanks
Bill

 

[rwooduk]

Excellent! Thanks for all the replies!

Very true.

But everyone into QM should have a smattering of knowledge of it:
http://physics.lamar.edu/rafa/cinvestav/second.pdf

Thanks
Bill

Very good link! Thank you!

 

[vanhees71]

More caveats can be found in the following very illuminating paper:

Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
http://arxiv.org/abs/quant-ph/9907069

 

[rwooduk]

More caveats can be found in the following very illuminating paper:

Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
http://arxiv.org/abs/quant-ph/9907069

Very informative, thanks!

I also found this useful to understand the basics:

http://www.eng.fsu.edu/~dommelen/quantum/style_a/herm.html

 

[rwooduk]

rather than start a new thread could I throw in a quick question about domains and ∈,

What is the domain of an observable? Is it simple the place when it can be seen? and also for this domain:

D(Q) = {f∈L^2|xf∈L^2}

I understand Q is the position operator, this this is where its position will reside, HOWEVER, what does f∈L^2 actually mean? Wiki says f would be an element of the set L^2, what's a set in this instance?

Thanks again

 

[dextercioby]

This is really abstract and the chosen example is a little tricky. D(Q) is the (maximal) domain of the operator, the set of all states for which you can measure the position of a particle. As for <f(x) is an element of L^(2)>, well it suffices to say that |f|² is Lebesgue integrable over a certain open domain of R³.

 

[Doug Huffman]

Leonard Susskind at Stanford U.'s YouTube channel has marvellous introduction and reviews from basic mechanics to advanced QM and cosmology.

 

[rwooduk]

This is really abstract and the chosen example is a little tricky. D(Q) is the (maximal) domain of the operator, the set of all states for which you can measure the position of a particle.

ahh i think i see now, it is the limit of where the operator can operate.

As for <f(x) is an element of L^(2)>, well it suffices to say that |f|² is Lebesgue integrable over a certain open domain of R³.

no idea what that means but it gives me something to look into, thanks for your reply!

Leonard Susskind at Stanford U.'s YouTube channel has marvellous introduction and reviews from basic mechanics to advanced QM and cosmology.

Yes, I have reached Lecture 5, I agree it's very good. Thanks.

 

[moriheru]

I would recommend Leonard Susskind theoretical minimum series he covers the basics of q bits and twospin system and derives all the basics equations such as Heisenbergs uncertainty principle and Schrödinger time(in)dependent equation from then on it's just harmonic oscillators and so on...

 

[rwooduk]

I would recommend Leonard Susskind theoretical minimum series he covers the basics of q bits and twospin system and derives all the basics equations such as Heisenbergs uncertainty principle and Schrödinger time(in)dependent equation from then on it's just harmonic oscillators and so on...

I have ordered this book. thank you!

 

[moriheru]

Good it's really great, so lots of fun,if you have any questions on the stuff just ask me.