Question on bra vs ket notation

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SUMMARY

The discussion clarifies the distinction between bra vectors and ket vectors in quantum mechanics, specifically in the context of spin states. Both bra and ket vectors are utilized to compute probability amplitudes and are equivalent in finite-dimensional vector spaces. The dual space of bras is isomorphic to the space of kets, yet they represent two distinct vector spaces conceptually. The Hilbert space allows for the identification of its dual space with itself, although this does not apply to generalized eigenvectors of self-adjoint operators in continuous spectra.

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Ahmed1029
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What's the difference between a bra vector and ket vector in specifying spin states except for notational convenience when calculating probablility amplitudes? Are they equivalent?
 
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Ahmed1029 said:
What's the difference between a bra vector and ket vector in specifying spin states except for notational convenience when calculating probablility amplitudes? Are they equivalent?

In the case of spin states, there answers is that there is no difference. This is because we are dealing with finite-dimensional vector spaces.
 
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andresB said:
In the case of spin states, there answers is that there is no difference. This is because we are dealing with finite-dimensional vector spaces.
The dual space (bras) is isomorphic to the space of kets. But, conceptually and notationally there are still two vector spaces here. Especially if we are using Dirac notation.
 
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PeroK said:
The dual space (bras) is isomorphic to the space of kets. But, conceptually and notationally there are still two vector spaces here. Especially if we are using Dirac notation.
Well, yes. Still, you can use one or the other space to compute probabilities and expected values in a very similar way, and that was the op question.
 
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The Hilbert space has the property that its dual space can be canonically identified with the Hilbert space itself. I.e., a given bound linear form ##L## is uniquely determined by a vector ##|L \rangle## via
$$L(|\psi \rangle)=\langle L|\psi \rangle.$$
Note that this does not (!) apply to generalized eigenvectors of a self-adjoint operator in the continuous part of its spectrum. Those refer to the dual of a dense subspace of the Hilbert space, where such an unbound self-adjoint operator, is defined, and which is larger than the Hilbert space. This becomes most clear in the "rigged-Hilbert-space formalism". For a short introduction, see, e.g., Ballentine, Quantum Mechanics.
 
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Ahmed1029 said:
What's the difference between a bra vector and ket vector in specifying spin states except for notational convenience when calculating probablility amplitudes? Are they equivalent?
Expressed in a fixed basis, the bra is a row vextor and the ket is a column vector.
A bra and a ket with the same label are conjugate transposed to each other; in particular, they need not contain the same numerical entries.
 
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