# The Unapologetic Mathematician

## Matrices 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, we’ve got an inner product space. That is, a vector space $V$, equipped with a choice of a particular inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$.

Now, any linear transformation $B:V\rightarrow V$ gives us a bilinear form. In our new notation we can write it as $B(v,w)=\langle v\rvert B\lvert w\rangle$. Given a basis $\left\{\lvert i\rangle\right\}_{i=1}^n$ we can write down the matrix $b_{ij}=\langle i\rvert B\lvert j\rangle$. Then if we’re given vectors $\langle v\rvert=v^i\langle i\rvert$ (notice how the Dirac notation can be rather context-dependent) and $\lvert w\rangle=\lvert j\rangle w^j$, we can put them together with $B$ to find

\displaystyle\begin{aligned}\langle v\rvert B\lvert w\rangle&=v^i\langle i\rvert B\lvert j\rangle w^j\\&=v^ib_{ij}w^j\end{aligned}

So this is indeed the same old matrix of the form.

We can read a lot of information about the form off of its matrix. As we proceed we’ll illustrate these various properties of bilinear forms, using the Dirac notation to (hopefully) make the ideas clearer.

July 8, 2009 Posted by | Algebra, Linear Algebra | 4 Comments

## Dirac Notation III

So we’ve got Dirac notation and it’s nice for inner product spaces, but remember we’re not just interested in vectors and vector spaces. We’re even more interested in transformations between vector spaces. So what happens when we throw a linear transformation into the mix?

To keep things simple, let’s just think of one vector space $V$ and a linear transformation $T:V\rightarrow V$. If we have a vector $v$, we can write it as the ket $\lvert v\rangle$. Then we can hit it with our transformation to get $Tv$. We can put this new vector into a ket as well: $\lvert Tv\rangle$. But we can also just write the transformation outside the ket, giving $T\lvert v\rangle$. Both of these notations make pretty good sense.

But what about bras? We can take the ket $\lvert Tv\rangle$ and flip it over into the bra $\langle Tv\rvert$. Now there’s no reason to believe that $T$ acts in exactly the same way on bra vectors as it does on ket vectors. So what does it do? Well, we can probe the bra $\langle Tv\rvert$ by seeing what happens when we feed it various kets. We pick a ket $\lvert w\rangle$ and pair up to form the bra-ket $\langle Tv\vert w\rangle$.

Now this is an inner product, and so we have the adjoint property, allowing us to rewrite it as $\langle v\vert T^*w\rangle$. This is the pairing of the bra $\langle v\rvert$ and the ket $\lvert T^*w\rangle$. But this latter ket we can also write as $T^*\lvert w\rangle$. So we could reasonably write out the pairing as $\langle v\rvert T^*\lvert w\rangle$. But we could then reasonably read this as the pairing of the ket $\lvert w\rangle$ with the bra $\langle v\rvert T^*$.

That is, when we have a transformation $T$ acting on a bra vector $\langle v\rvert$ to give the bra $\langle Tv\rvert$, we can see this as the adjoint of $T$ acting from the right in our notation. Then expressions like $\langle v\rvert T\lvert w\rangle$ can be interpreted either as $T$ acting on $w$ and then paired with $v$, or as $T^*$ acting on $v$, which then pairs with $w$. And this is actually no ambiguity, since the way adjoints and inner products interact guarantees that these both give the same answer in the end.

Incidentally, what is an expression like $\langle v\rvert T\lvert w\rangle$ anyway? It’s a matrix element of the transformation $T$. In fact, if we have a basis $\left\{\lvert i\rangle\right\}_{i=1}^n$ then the matrix components can be written in Dirac notation as $t_j^i=\langle i\rvert T\lvert j\rangle$. The Dirac notation makes clear the way that bra and ket vectors probe two different aspects of linear transformations.

In quantum physics, we see exactly this sort of probing going on all the time. Except that there, the physicsts talk about picking an “input state” $\lvert w\rangle$ and an “output state” $\langle v\rvert$, and thinking of a physical process as a linear transformation $T$ from a space of input states to a space of output states. The “transition amplitude” — connected with the probability of ending up at the given output when starting at the given input — is then given by $\langle v\rvert T\lvert w\rangle$. It’s just a matrix element of the linear transformation representing out physical process.

July 6, 2009 Posted by | Algebra, Linear Algebra | 3 Comments

## Dirac 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 dual, with the ket $\lvert v\rangle$ denoting a vector $v\in V$ and the bra $\langle\lambda\rvert$ denoting a linear functional $\lambda\in V^*$. The evaluation $\lambda(v)$ is then denoted by the bra-ket pairing $\langle\lambda\vert v\rangle$.

But the neat thing about this notation is that it makes bras look like some sort of reflection of kets. And they are, in a sense. The dual space $V^*$ is some sort of reflection of the vector space $V$, but there’s no clear mapping from vectors in one space to vectors in the other; unless, that is, we pick a specific isomorphism; or, equivalently, an inner product.

When we’ve got an inner product in the picture, we get a (conjugate) linear isomorphism that sends the vector $v$ to the linear functional $\langle v,\underline{\hphantom{X}}\rangle$. In Dirac notation, we send the ket $\lvert v\rangle$ to the bra $\langle v\rvert$. Then the value of this linear functional on a vector $w$ (the ket $\lvert w\rangle$) is the pairing $\langle v\vert w\rangle=\langle v,w\rangle$, just as it should be.

July 1, 2009 Posted by | Algebra, Linear Algebra | 6 Comments