Bra Vectors

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One question that's been on my mind for a while is what the difference between a bra vector and a covector is. Are they the same thing? Why the difference of notation?
 

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  • #2
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Where did you encounter co-vectors in quantum theory? And what did they look like?

Cheers,

Jazz
 
  • #3
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Where did you encounter co-vectors in quantum theory? And what did they look like?

Cheers,

Jazz
I didn't. I'm a math guy. :tongue:

I read a thing on how the Riemann Hypothesis is related to quantum physics, and the article used some bras and kets. I assumed, from context, that a bra vector was essentially the same as a covector. I wanted to know if I was right.
 
  • #4
WannabeNewton
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There is no difference. A bra is a covector but you should keep in mind that the term dual vector is more appropriate in the general linear algebra setting; the term covector is more prevalent in differential geometry. The notation is more convenient in the context of QM is all.
 
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  • #5
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Well, you said they differed in notation, so I was surprised you found a different notation alongside the Dirac notation. And yes, bras are the same concept as co-vectors, both are the dual with respect to the inner product. But you wouldn't call the bra a co-vector usually. The term co-vector is used in the context of manifolds in physics, so typically either symplectic geometry (hamilton formalism) or riemannian geometry (GR).

Hope this clears it up :)

Cheers,

Jazz
 
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  • #6
dextercioby
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One question that's been on my mind for a while is what the difference between a bra vector and a covector is. Are they the same thing? Why the difference of notation?
A co-vector is a more general notion than of a bra-vector, because you needn't have a topology, nor a scalar product to speak about vectors and co-vectors, but you need to have them to speak of bra-s and ket-s.
 
  • #7
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A co-vector is a more general notion than of a bra-vector, because you needn't have a topology, nor a scalar product to speak about vectors and co-vectors, but you need to have them to speak of bra-s and ket-s.
That's interesting you bring that up. There seem to be two different concepts or definitions of co-vectors in the literature, depending on whether you get there using the exterior algebra and differential forms or multilinear forms and a metric (with the metric as the defining bijection for the dual). For the diff-forms approach you only need the metric when you introduce the hodge dual and not the co-vectors.

I think I understand this all pretty well, but I never found the different definitions very intuitive and the differences rather confusing. But then again, I'm only a theoretical physicist and not a real mathematician.

Any thoughts on that?

Cheers,

Jazz
 
  • #8
dextercioby
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I think 'co-vectors' come uniquely from linear algebra. They are automatically there when you have a vector space. But the concept of vector space is useful also in geometry and that's how we bring 'co-vectors' in geometry. There needn't be a metric on a manifold, nor a connection. Co-vectors are there because vectors are there. That's always the case.
 
  • #9
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I think 'co-vectors' come uniquely from linear algebra. They are automatically there when you have a vector space. But the concept of vector space is useful also in geometry and that's how we bring 'co-vectors' in geometry. There needn't be a metric on a manifold, nor a connection. Co-vectors are there because vectors are there. That's always the case.
So what's your most general definition of a co-vector?

Cheers,

Jazz
 
  • #10
dextercioby
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If a is a member of a vector space V (called 'vector') over the field K, then b taking a into a unique element of K is a co-vector, or a (linear) functional over V.
 
  • #11
rubi
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The dual space ##V^*## of a topological vector space ##V## (over ##\mathbb C##) is the space of continuous linear functionals ##f:V\rightarrow\mathbb C##. A member of ##V^*## is called dual vector or co-vector. This is mostly useful if ##V## is locally convex (e.g. Banach spaces or Hilbert spaces, the Schwarz space, ...), since then the Hahn-Banach theorem guarantees the existence of enough such functionals to make the theory interesting. On a Hilbert space ##H## you have Riesz's theorem, which tells you that ##H^*## is isomorphic to ##H## (this is used a lot in QM).

In the case of manifolds, you automatically have the tangent spaces ##T_p M## at every point and since they are topological vector spaces (all finite-dimensional vector spaces have this property), you can form their dual ##T^*_p M##. If you have a metric, the tangent spaces are Hilbert spaces and you can use the Riesz isomorphism to identify tangent vectors with cotangent vectors.
 

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