# Hermitian momentum operator ?

1. Mar 8, 2014

### dyn

A Hermitian operator A is defined by A=A(dagger) which is the transpose and complex conjugate of A. In 1-D the momentum operator is -i(h bar)d/dx. How can this be Hermitian as the conjugate has the opposite sign ?
Thanks

2. Mar 8, 2014

### strangerep

For unbounded operators with infinite spectrum, such as the momentum operator, one should use the adjective "self-adjoint" instead of Hermitian. If $\phi, \psi$ are square-integrable wave-functions, then a linear operator $A$ is said to be self-adjoint if
$$\int\!dx\; \phi^*(x) \, A \psi(x) ~=~ \int\!dx \Big(A \phi(x)\Big)^* \, \psi(x) ~.$$Exercise: check whether the above holds when $A$ is the momentum operator.
(Hint: use integration by parts.)

3. Mar 8, 2014

### dyn

So the momentum operator isn't Hermitian ? But it still returns real eigenvalues and expectation values. Does it also produce orthogonal wavefunctions ?

4. Mar 8, 2014

### The_Duck

Checking whether $A = A^\dagger$ isn't quite as easy as it looks: the derivative operator $\frac{d}{dx}$ is in fact anti-Hermitian, so that the momentum operator as a whole is Hermitian (unsophisticated people like myself treat "Hermitian" and "self-adjoint" as synonyms, though in reality as strangerep points out this ignores certain subtleties).

"$A = A^\dagger$" is basically shorthand for the expression strangerep wrote. You should check whether that expression holds when $A$ is the momentum operator. If it holds, then the momentum operator is Hermitian.

5. Mar 8, 2014

### strangerep

The notion of self-adjointness is a generalization of the notion of Hermiticity. E.g., in a finite dimensional vector, where $u,v$ are vectors (with complex components), the usual inner product can be written as
$$(u,v) ~:=~ u^\dagger v ~.$$ If $A$ is a matrix, it is self-adjoint if $(Au, v) = (u,Av)$ or in this case:$$u^\dagger A v ~=~ (Au)^\dagger v ~.$$Thus, self-adjointness coincides with Hermiticity in the finite-dimensional case, i.e., is equivalent to usual notion of Hermiticity as the transpose conjugate.

Expressing the notion in terms of the inner product allows it to be generalized to infinite-dimensional Hilbert spaces of wave functions, on which the momentum operator acts. In that case, Hermiticity and self-adjointness no longer coincide, in general. Self-adjointness still guarantees real eigenvalues (exercise).

As for orthogonality of eigenfunctions, the concept of eigenfunction must be generalized since those of the momentum operator are not normalizable, hence not in the Hilbert space. But that's a long story.

What textbook(s) are you studying from? I.e., QM and linear algebra? If you don't have any, then try Ballentine for QM and maybe Axler for linear algebra.

6. Mar 8, 2014

### micromass

Staff Emeritus
Isn't this hermitian? In order to be self-adjoint, you need some subtleties with the domain to hold.

7. Mar 8, 2014

### strangerep

Ssshh. I was trying to coax the OP away from the notion of transpose+conjugate, and think in terms of the inner product instead in order to generalize to inf-dim spaces. I had intended (perhaps) to mention domains later, but let's wait and see what textbook(s) the OP is using, and what level of answer is appropriate...

8. Mar 8, 2014

### micromass

Staff Emeritus
Fair enough :tongue: Just found it weird that you corrected hermitian by self-adjoint but then write the condition for hermitian anyway. But yeah, let's wait for the OP to come back.

9. Mar 8, 2014

### strangerep

I'll rethink my pedagogical technique in future.

10. Mar 8, 2014

### micromass

Staff Emeritus
Don't, I'm just nitpicking :shy:

11. Mar 8, 2014

### dyn

I am mainly studying from QM textbooks. They all seem to say that the momentum operator is Hermitian and I can see that it satisfies the equations you have mentioned but I couldn't see how it satisfied A=A(dagger).

12. Mar 8, 2014

### micromass

Staff Emeritus
Take a look at strangerep's post #2. Do you agree that you need to check that relation for $A$? Why don't you plug in your $A$ in that integral and work it out?

13. Mar 8, 2014

### strangerep

I agree with Micromass. It's time for you to do some work. (You couldn't even be bothered to say which textbooks.)

14. Mar 9, 2014

### stevendaryl

Staff Emeritus
As others have said, the way to understand this is by looking at the definition of $A^\dagger$ for inner products, which means an integral.

For the special case of $A = \frac{d}{dx}$, we have:

$\int \Psi^*(x) \frac{d}{dx} \Phi(x) dx = \int \frac{d}{dx}(\Psi^*(x) \Phi(x)) dx - \int (\frac{d}{dx} \Psi^*(x)) \Phi(x) dx$

That's just using the product rule for derivatives. The first term on the right vanishes for sufficiently well-behaved functions $\Psi$ and $\Phi$, when we integrate over the entire space, so we have:

$\int \Psi^*(x) \frac{d}{dx} \Phi(x) dx = - \int (\frac{d}{dx} \Psi^*(x)) \Phi(x) dx$

That minus sign on the right is what makes $\frac{d}{dx}$ anti-hermitian.

15. Mar 9, 2014

### dyn

I am using a variety of textbooks including Shankar , Rae , Griffiths , Mandl and Zettilli