# The Unapologetic Mathematician

## Dirac notation I

There’s a really neat notation for inner product spaces invented by Paul Dirac that physicists use all the time. It really brings to the fore the way a both slots of the inner product enter on an equal footing.

First, we have a bracket, which brings together two vectors

$\displaystyle\langle w,v\rangle$

the two sides of the product are almost the same, except that the first slot is antilinear — it takes the complex conjugate of scalar multiples — while the second one is linear. Still, we’ve got one antilinear vector variable, and one linear vector variable, and when we bring them together we get a scalar. The first change we’ll make is just to tweak that comma a bit

$\displaystyle\langle w\vert v\rangle$

Now it doesn’t look as much like a list of variables, but it suggests we pry this bracket apart at the seam

$\displaystyle\langle w\rvert\lvert v\rangle$

We’ve broken up the bracket into a “bra-ket”, composed of a “ket” vector $\lvert v\rangle$ and a “bra” dual vector $\langle w\rvert$ (pause here to let the giggling subside) (seriously, I taught this to middle- and high-schoolers once).

In this notation, we write vectors in $V$ as kets, with some signifier inside the ket symbol. Often this might be the name of the vector, as in $\lvert v\rangle$, but it can be anything that sufficiently identifies the vector. One common choice is to specify a basis that we would usually write $\left\{e_i\right\}$. But the index is sufficient to identify a basis vector, so we might write $\lvert1\rangle$, $\lvert2\rangle$, $\lvert3\rangle$, and so on to denote basis vectors. That is, $\lvert i\rangle=e_i$. We can even extend this idea into tensor products as follows

$\displaystyle e_i\otimes e_j=\lvert i\rangle\otimes\lvert j\rangle=\lvert i,j\rangle$

Just put a list of indices inside the ket, and read it as the tensor product of a list of basis vectors.

Bras work the same way — put anything inside them you want (all right, class…) as long as it specifies a vector. The difference is that the bra $\langle w\rvert$ denotes a vector in the dual space $V^*$. For example, given a basis for $V$, we may write $\langle i\rvert=\epsilon^i$ for a dual basis vector.

Putting a bra and a ket together means the same as evaluating the linear functional specified by the bra at the vector specified by the ket. Or we could remember that we can consider any vector in $V$ to be a linear functional on $V^*$, and read the bra-ket as an evaluation that way. The nice part about Dirac notation is that it doesn’t really privilege either viewpoint — both the bra and the ket enter on an equal footing.

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Posted by John Armstrong | Algebra, Linear Algebra

## 14 Comments	»

1. But why not simply consider the linear map ev: V* (x) V -> k, where k is the ground field, using simply the notation (a,b) for the nondegenerate pairing (if a \in V, b \in V*)?

Comment by Zygmund | June 30, 2009 | Reply

2. Trivia: “bra” means good in Swedish.

Comment by Å | June 30, 2009 | Reply

3. Å: good to know

Zygmund: there are other things we’re going to want to do, especially if we’ve got an inner product around (like in a Hilbert space for quantum mechanics). Stay tuned.

Comment by | June 30, 2009 | Reply

4. Thank you, John Armstrong. I’m hard at work redlining Draft 4.0 of a QM paper. Question is: what is the topology of the ensemble of theories of QM if we don’t presume that Planck’s contant is a nonnegative real number? You clarify some notational issues that are deeper than the look.

Comment by | June 30, 2009 | Reply

5. Naive question: if putting a bra and ket together does it what it does for two vectors, what is the equivalent for a “3-bra-ket” widget that operates on three vectors? Does it map to a vector triple products, or is there a Jacobi identity, or what?

Comment by | June 30, 2009 | Reply

6. As far as I know there’s no natural three-variable operation analogous to the pairing between a vector space and its dual.

Comment by | June 30, 2009 | Reply

7. Actually, that’s not quite true. In special cases there is such an operation. One in particular is the “triality” between the three eight-dimensional irreducible representations of $\mathfrak{so}(8)$.

Comment by | June 30, 2009 | Reply

8. Thank you, John Armstrong, for that fascinating comment #7. I can’t even claim that fact was “on the tip of my tongue.” But I had a nagging 1/3 memory that there was something beyond duality. This comes from, I think, but please correct me, special features of the group Spin(8), i.e. the double cover of the 8-dimensional rotation group SO(8), arising because the group has an outer automorphism of order three? Wasn’t this discussed on the n-category Cafe or John Baez’s blog?

Comment by | June 30, 2009 | Reply

9. Yes, Baez’ has discussed it in “This Week’s Finds”.

Comment by | June 30, 2009 | Reply

10. [...] Notation II We continue discussing Dirac notation by bringing up the inner product. To this point, our notation applies to any vector space and its [...]

Pingback by | July 1, 2009 | Reply

11. The post says that, for <w, v>, “the first slot is antilinear — it takes the complex conjugate of scalar multiples — while the second one is linear”. This seems to be the opposite of what Wikipedia says for inner product space:
http://en.wikipedia.org/wiki/Inner_product_space#Definition

What am I missing?

Comment by Sig Freud | July 5, 2009 | Reply

12. You’re missing the fact that which slot is which is a choice of convention, and I’ve chosen the other convention than they have. Go back and read my earlier posts.

Comment by | July 5, 2009 | Reply

13. [...] Notation III So we’ve got Dirac notation and it’s nice for inner product spaces, but remember we’re not just interested in [...]

Pingback by | July 6, 2009 | Reply

14. [...] and Bilinear Forms on Inner Product Spaces in Dirac Notation Now, armed with Dirac notation, we can come back and reconsider matrices and forms. For our background, [...]

Pingback by | July 8, 2009 | Reply

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