# Math Help - inner product definition on C

1. ## inner product definition on C

There seems to be two definitions for the inner product <f|g> used in Hilbert space: $\int$ $\overline{f(t)}$g(t)dt, and
$\int$f(t) $\overline{g(t)}$dt.
I understand that if one is consistent, either definition will work. Nonetheless, which is (more) standard?

2. The first one is, in my opinion, much the superior version. It's the one the physicists use, which means you get to employ the power of the Dirac formalism, which, I see, you have employed there. If you use the notation <f|g>, then you mean the first one. I've never seen a mathematician use the Dirac notation, and also mean the second definition.

3. Ackbeet's answer is absolutely spot-on. Most mathematicians use the second definition, and most (all?) physicists use the first definition. But some mathematicians are moving over to the physicists' version, which has one big advantage. If you use the tensor product notation to denote rank-one operators, then the operator $x\mathord{\scriptstyle\otimes} y$ is defined (according to the physicists' definition for the inner product) by the very natural-looking formula $(x\mathord{\scriptstyle\otimes} y)z = x\langle y,z\rangle$. But if you use the mathematicians' definition of the inner product then the tensor product operator has to be defined as $(x\mathord{\scriptstyle\otimes} y)z = \langle z,y\rangle x$, which means that you are constantly having to juggle the order of the elements x, y, z.

(In fact, the tensor product formalism is exactly the same as the Dirac formalism, using a different notation.)

4. Incidentally: a nice, prettified LaTeX version of the definition would look like the following. The inner product on $\mathcal{L}^{2}(A)$ is defined by

$\displaystyle\langle f|g\rangle=\int_{A}\overline{f(x)}\,g(x)\,dx.$

5. Thanks very much, Ackbeet and Opalg.

6. (deleted)

7. Just to chime in on the Physicist's side of all this, the Dirac (<f|g>) notation is more commonly used in advanced Quantum Mechanics where we rarely consider the actual wavefunction. But the integral form is much easier to apply when actually calculating a wavefunction. Though it is, of course, possible to use either definition.

-Dan