Bra & Ket notation in quantum physics

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Homework Help Overview

The discussion revolves around the application of bra and ket notation in quantum physics, specifically focusing on the inner product of a state vector |A> expressed as a linear combination of two orthonormal states |B> and |C>.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the expansion of the inner product using the properties of orthonormal states. Questions arise regarding the definitions and implications of orthonormality versus orthogonality.

Discussion Status

Participants are actively engaging with the problem, providing guidance on how to expand the expression and simplify it. There is a recognition of the properties of orthonormal systems, and some clarification on terminology has been achieved.

Contextual Notes

There is a mention of potential confusion between orthonormal and orthogonal states, indicating a need for clarity in definitions as part of the discussion.

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Homework Statement



If |A> = x |B> + y |C> where |B> and |C> are orthonormal, then what happens when <A|A> ?

The Attempt at a Solution



Would <A| = x* <B| + y* <C|?

I'm not really sure where to go from there
 
Last edited:
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hi blueink! welcome to pf! :smile:

now expand (x* <B| + y* <C|)(x |B> + y |C>) :wink:

[and what is <B||B>? what is <B||C>?]
 
I'm not really sure where to go from there
Write down <A|A> with your <A|, and use the distributive property of the scalar product:
<X| (|Y>+|Z>) = <X|Y> + <X|Z>
Afterwards, simplify.
 
Expanding (x* <B| + y* <C|)(x |B> + y |C>)
=(x* <B|)(x |B>)+(x* <B|)(y |C>)+(y* <C|)(x |B>)+(y* <C|)(y |C>)
=x*x <B|B> + x*y <B|C> + y*x <C|B> + y*y<C|C>

Where <B|B> and <C|C> = 1 and <B|C> and <C|B> = 0?
 
yup! :smile:

that's what's so beautiful about orthonormal systems! :wink:

(erm … you did mean orthonormal, and not just orthogonal? :redface:)
 
tiny-tim said:
yup! :smile:

that's what's so beautiful about orthonormal systems! :wink:

(erm … you did mean orthonormal, and not just orthogonal? :redface:)

I did mean orthonormal, ahh confusion! thanks :)
 

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