# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Algebra, Linear Algebra

## 3 Comments	»

1. I’m really enjoying this section Dirac notation. If I explain this to students, though, I might have to avoid using the phrase, “…we can probe the bra…”
Many thanks for the fine explanations.

Comment by | July 6, 2009 | Reply

2. [...] 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 [...]

Pingback by | July 8, 2009 | Reply

3. [...] First, we hit each of these vectors with . The ket vector clearly becomes , while the bra vector becomes . Now using the form from before we [...]

Pingback by | July 24, 2009 | Reply

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