1. Mar 20, 2007

### pivoxa15

Is the adjoint of an opertor equivalent to the complex conjugate of that operator?

So if the operator is A then the adjoint of A is A* where * denotes the complex conjugation.

So A is self adjoint <=> A=A* or A= adjoint of A

2. Mar 20, 2007

### mjsd

adjoint usually means complex conjugate transpose

3. Mar 20, 2007

### HallsofIvy

Staff Emeritus
What do YOU mean by "the complex conjugate" of an operator?

4. Mar 20, 2007

### matt grime

If H is a hilbert space, and L in End(H), then the adjoint, K, of L is the element in End(H) that satisfies

<Lx,y>=<x,Ky>

for all x,y. From the definition, it is not clear that such a K ever exists, but it does (Rietsche representation lemma - or abstract nonsense from category theory) for Hilbert spaces.

If we pick basis, and happen to have a finite dimensional hilbert space, so that we identify L with a matrix M(L), then M(K) is the conjugate transpose of M(L).

5. Mar 20, 2007

### pivoxa15

I equally would like to know what is the adjoint (usually portrayed by a dagger sign) of an operator A?

Wahtever, it is Could you say that A (dagger)=A* where * denotes complex conjugation.

6. Mar 20, 2007

### matt grime

I told you what the adjoint of an operator is. You can't 'complex conjugate' an arbitrary operator, since they are not collections of complex numbers. That is Halls's point.

7. Mar 21, 2007

### pivoxa15

I see. A Hermition operator is where the adjoint of the operator = the operator itself. A^=A where ^ denotes transpose.

A is hermition <=> <u,A^v>=<u,Av>

OR <A(v),w>=<v,A(w)>

Althogh what is <A(v),w>*=?

The reason why I mentioned conjugate is because I was thinking about the specific case of operators in QM. However in there, u and v are complex conjugates so a Hermition operator also means that the expectation values are real. So the expectation value = the complex conjugate of the expectation value.

Last edited: Mar 21, 2007
8. Mar 22, 2007

### matt grime

No, it is where A=A^, and A^ means the adjoint of A, which is not the transpose. Just as it doesn't make sense to talk of the conjugate of an operator per se, it doesn't make sense to talk of the transpose of an operator. Stop assuming these things are matrices!

d/dx is a prefectly good operator on certain hilbert spaces. What is the transpose of the differentiation operator?

The adjoint of A satisfies <Av,w>=<v,A^w> for all v,w, so the second of those options is correct.

9. Mar 22, 2007

### dextercioby

There's certain delicacy that must be used once one deals with infinite dimensional Hilbert spaces. One of them occurs when trying to define the adjoint of a linear (not necessarily densly defined, nor bounded) operator. Sometimes it doesn't exist as a linear operator. Even when it does, it might turn out that its domain could only be the null vector...

10. Mar 22, 2007

### pivoxa15

I see. Adjoint is the universal language applied to operators or matrices or anything else.

If we use this definition <Av,w>=<v,A^w> and apply to QM then

<p*,A^p> = <(Ap)*,p> where <> denotes integration and comma is multiplication. p = psi wave function. * is complex conjugation. Does this match with the general definition? If v=p* then it should be <p*,A^p>=<A(p*),p> which is different to what I gave or is it the same?

Last edited: Mar 22, 2007
11. Mar 23, 2007

### matt grime

That is not what < , > means in QM.

<p,q> just means an (complex) iner product, and for hilbert spaces that you mention there the innerproduct is

$$\int p*(x)q(x)dx$$

But < > does not mean integrate, and , does not mean multiply, and you have the * in the wrong place.

12. Mar 24, 2007

### pivoxa15

So the hilbert space is an inner product space. Does that mean all the linear algebra in inner product space I have learnt since 1st year are actually hilbert spaces? I use to think that hilbert spaces were very specialised spaces.

Things seem to work out after your suggestion.

13. Mar 24, 2007

### matt grime

yes. but one with an extra assumption - that is complete as a normed space.

no. but it all applies to a hilbert space, since hilbert spaces are a subclass of inner product spaces. how on earth did they define a hilbert space for you?

the only special thing about them is that they are complete as normed spaces.

14. Mar 24, 2007

### pivoxa15

They never mentioned hilbert spaces but mathematics in hilbert spaces looks very similar to mathematics in any inner product space I have done.

Does complete as normed spaces mean that a metric is defined in hilbert space as a norm (i.e. ||.||) as shown here http://en.wikipedia.org/wiki/Hilbert_space.
We were introduced to this norm in 1st year and were told that was how to calculate length in inner product spaces. Dosen't that mean they were really talking about hilbert spaces?

15. Mar 24, 2007

### matt grime

For the 3rd time, a hilbert space is a complete normed innerproduct space. So R^n, C^n are hilbert spaces with the usual norm. Q^n is an inner product space in the obvious way but is not a hilbert space. To repeat that I'm repeating myself, Hilbert spaces are a special case of an inner product space. It's as if you're saying - they never mentioned the real numbers were a group, so when they were talking about groups they were really talking about the real numbers. No!