# Dirac notation expressions as integrals

1. Aug 12, 2012

### maalpu

Can anyone point me to how to interpret Dirac notation expressions as wave functions and integrals beyond the basics of
<α| = a*(q)
|β> = b(q)
<α|β> = ∫ a* b dq

For example in the abstract Dirac notation the expression
|ɣ> (<α|β>)
can be evaluated as
(|ɣ><α|) |β>
Ω |β>
|ω>
but what can you do with the equivalent integral
g ∫ a* b dq
to combine g and a on the way to a final function w based on b ?

And what does it mean for an operator to operate on a function to the left - if it is simply that
f O = O* f
then O f* is also possible, yet
Ω <φ|
is not permitted in the abstract notation ?

2. Aug 12, 2012

### Chopin

Just use different integration variables.

$$|g\rangle \langle a | b \rangle = g(x) \int{a^*(y)b(y) dy} = \int{\Big[g(x)a^*(y)\Big] b(y) dy}$$
So you can interpret $\Omega = |g\rangle \langle a |$ as $\Omega(x,y) = g(x)a^*(y)$. Then, when you apply that operator to a state $|b\rangle$, you get what you would expect:
$$|\Omega|b\rangle = \int{\Omega(x,y)b(y)dy} = \int{g(x)a^*(y)b(y) dy}$$
Just like above, the notion of applying an operator to a state in this notation really just means matching up integration variables. It doesn't matter which side the operator is on, since in this notation they're just plain old numbers. So we have:
$$\langle \psi | \Omega = \int \psi^*(x)\Omega(x,y) dx = \int \Omega(x,y) \psi^*(x) dx$$
The fact that we put the bra on the left in Dirac notation is just a way of ensuring that things match up properly when we combine bras and kets with operators.

Last edited: Aug 12, 2012
3. Aug 14, 2012

### maalpu

Thanks Chopin - I had not thought about or seen that emphasis on the integration variables.

The rule then seems be that bra-ket has common integration variables, and ket-bra different ones, so <f|g><a|b> becomes f*(x) g(x) a*(y) b(y) whichever way they are grouped.

For the operator, here it was just a simple number, a multiplier, but in general it might be eg a partial differential - can it then still be simply applied either left or right ?

And I notice with |g> above you didn't introduce an ∫dx, but in the final <ψ|Ω you did (and it is by definition just some other bra <ξ|), an inconsistency I see in many places - when does a bra or ket outside a bra-ket imply integration and when not ?

4. Aug 14, 2012

### tom.stoer

For any abstract state |ψ> and a (continuous) basis |a> defined as eigenstates of some observable A you may define

$$1 = \int da\,|a\rangle\langle a|$$

Then you can insert this identity and you'll get

$$|\psi\rangle = \int da\,|a\rangle\langle a|\psi\rangle = \int da\,\psi(a)\,|a\rangle$$

with the "wave function" as projections w.r.t. to |a>

$$\psi(a) = \langle a|\psi\rangle$$

hope this helps

5. Aug 14, 2012

### Chopin

Correct, except that you also need $\int{dx\:dy}$ in there for it to make sense, but I think you probably got that already.

In that case, you pretty much have to keep the operator to the left of the quantity, but that's just because our rules for writing down derivatives say that the $\partial$ always applies to the thing to the right of it. This $\Omega(x,y)$ notation for operators doesn't really handle derivatives very well--you can do it, but it involves a bunch of funny Fourier transforms that make it look a lot more complicated than it really is.

Basically, any time you have a $|$ in an expression with something on both sides of it, you will integrate over a common variable (I think Feynman even once said that the great rule of quantum mechanics is simply that $| = \int$). So $|g\rangle = g(x)$ doesn't have one, because there's nothing to integrate it with, but $\langle \psi|\Omega = \int{\psi^*(x)\Omega(x,y)dx}$ does, because there are two things to multiply together.

Have you taken a course in linear algebra? If so, then this may be familiar to you in other terms. Technically speaking, when we put together a bra and a ket, what we're really doing is taking the inner product of two vectors. In a normal finite-dimensional vector space, taking an inner product can be done by breaking the vectors apart into a common basis, multiplying each component with the other, and adding them all together (i.e. the dot product). Doing an integral is just the continuous equivalent of this, where the basis vectors are just a continuum of position eigenstates, and $g(x)$ just tells us the weight of each one in the vector. Similarly, an operator is technically a rank-2 tensor that we take the product of with other vectors. That's why it has two different integration variables instead of just one.

If you haven't taken linear algebra, that paragraph may not make much sense. But even if you're not familiar with what's technically going on, the key to remember is that any time you're sticking two things together in Dirac notation, you have to do an integration over their product in integral notation. Doing that will kill their common variable, and leave behind any other variables that might have been lying around, which you can then use to hook up with other functions later on.

Last edited: Aug 14, 2012
6. Aug 14, 2012

### Muphrid

I would get out of the habit of thinking of $| \psi \rangle \equiv \psi(x)$. The way I was taught was that only $\langle x | \psi \rangle = \psi(x)$. This makes what's going on clear: if you have a finite basis, you can always expand in terms of those states by a summation. That is,

$$| \psi \rangle = \sum_i | i \rangle \langle i | \psi \rangle$$

Similarly, on an infinte basis (like the position basis), the summation turns into an integral:

$$| \psi \rangle = \int_{-\infty}^\infty | x \rangle \langle x | \psi \rangle \; dx$$

You can choose the basis to expand on by whatever's convenient--position basis, wavenumber basis, whatever.

7. Aug 14, 2012

### tom.stoer

This is the key issue.

8. Aug 14, 2012

### Chopin

I'd agree as well that this way is better, albeit a bit more verbose--it makes it much more obvious which basis you're working in. Similarly, $\Omega(x,y) = \langle x|\Omega|y\rangle$, which is why $\Omega(x,y)$ are sometimes referred to as the matrix elements of $\Omega$--you can think of the elements forming a big two-dimensional matrix, just like you can think of a ket as forming a big column vector, and a bra as a big row vector.