1. Dec 6, 2007

### jostpuur

Is there any other difference between self-adjoint operators, and Hermitian operators, than that mathematicians seem to talk mostly about self-adjoint operators, and physicists seem to talk mostly about Hermitian operators?

2. Dec 6, 2007

### HallsofIvy

Staff Emeritus
Mathematicians tend to talk more abstractly than physicists!

The Hermitian (more correctly, the Hermitian adjoint) of an operator only apply to operators on vector spaces over the complex numbers.

If U and V are any inner product spaces and T is a linear transformation from U to V, the "adjoint" of T, T*, is a linear transformation from V to U such that, for any u in U and v in V, <Tu,v>= <u,T*v>. The two inner products are taken in V and U respectively.

In particular if U= V and T*= T, that is, if <Tu,v>= <u,Tv>, then T is "self-adjoint".

3. Dec 6, 2007

### jostpuur

And it is allowable to call T Hermitian only if U is a vector space over complex field?

4. Dec 6, 2007

### HallsofIvy

Staff Emeritus
I wouldn't be hard-nosed about it! Both mathematicians and physicists sometimes are loose with terminology. (I just checked two Linear Algebra texts to see exactly how they defined "Hermitian" and the word is not even in the index!)

5. Dec 6, 2007

### Galileo

IIRC correctly:
-An operator is Hermitian if <u,Av>=<Au,v> for all u,v, in the domain of A.
-An operator is self-adjoint if A*=A.

The subtle difference being that the domains of A and A* may not coincide in general.
These definitions coincide for bounded operators.
I've seen some places where a hermitian operator is bounded by definition, others relax that condition. I`m not too sure of the exact definitions so don't quote me here.

6. Feb 29, 2008

### jostpuur

I had difficulty understanding this because I knew only one definition for A* which was for bounded A only (through the Riesz's representation theorem). I've just learned that for unbounded operators A, A* can still be defined, but differently. I think I got this now.

Last edited: Mar 1, 2008
7. Feb 29, 2008

### MaWM

Correct. But it's more than that. A Hermitian operater is a self adjoint operator on a complex inner product space with a particular inner product. So, even if the field is complex, if the inner product is is the typical one, the operater would be merely selfadjoint.

For a matrix, the Hermitian conjugate is the complex conjugate of the transpose. If this is equal to the original, the matrix is Hermitian.

Last edited: Feb 29, 2008
8. Mar 1, 2008

### jostpuur

This was very confusing thread. I'll write the definition of the adjoint that I have just encountered.

Let T:D(T)->H be an operator, possibly unbounded, so that D(T) is dense in H. We then define

$$D(T^*) = \{ x\in H\;|\; \underset{y\in D(t), \|y\|=1}{\textrm{sup}} |(x|Ty)| <\infty\},$$

and it is possible to define T*:D(T*)->H by setting (T*x|y) = (x|Ty) for all x in D(T*) and y in D(T).

I don't know the details of the proof needed for this definition yet, but this looks good anyway.

Considering Hallsoflvy's first post, I think my attention was drawn to this

too much, while I didn't know what was relevant. The rest of the post was already containing the answer, which was the same answer as given by Galileo... But I'm not convinced that everything was fine with the sets U and V here

With arbitrary norm spaces the adjoint would be between the duals, T*:V*->U*. With Hilbert spaces, I assume, it goes like I showed now.

Anyway, my difficulty rose from the fact that I only knew T->H and T*->H case earlier with bounded operators. Also, if a bounded operator T:D(T)->H is defined on a dense subset, it can always be extended to the H uniquely.

Looking back at the Galileo's answer, I should have been able to ask about the definition of the adjoint....... but you know, it's so difficult to keep the thoughts clear

Last edited: Mar 1, 2008