# Bra Vectors

1. Oct 5, 2013

### Mandelbroth

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?

2. Oct 5, 2013

### Jazzdude

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

Cheers,

Jazz

3. Oct 5, 2013

### Mandelbroth

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. Oct 5, 2013

### WannabeNewton

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.

5. Oct 5, 2013

### Jazzdude

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

6. Oct 5, 2013

### dextercioby

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. Oct 5, 2013

### Jazzdude

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. Oct 5, 2013

### dextercioby

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. Oct 5, 2013

### Jazzdude

So what's your most general definition of a co-vector?

Cheers,

Jazz

10. Oct 5, 2013

### dextercioby

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. Oct 5, 2013

### rubi

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.