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

Related Calculus and Beyond Homework Help News on Phys.org
mjsd
Homework Helper
adjoint usually means complex conjugate transpose

HallsofIvy
Homework Helper
What do YOU mean by "the complex conjugate" of an operator?

matt grime
Homework Helper
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).

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

matt grime
Homework Helper
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.

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:
matt grime
Homework Helper
I see. A Hermition operator is where the adjoint of the operator = the operator itself. A^=A where ^ denotes transpose.

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?

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

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

dextercioby
Homework Helper
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...

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.
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:
matt grime
Homework Helper
<p*,A^p> = <(Ap)*,p> where <> denotes integration and comma is multiplication. p = psi wave function. * is complex conjugation.
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.

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.
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.

matt grime
Homework Helper
So the hilbert space is an inner product space.
yes. but one with an extra assumption - that is complete as a normed space.

Does that mean all the linear algebra in inner product space I have learnt since 1st year are actually hilbert spaces?
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?

I use to think that hilbert spaces were very specialised spaces.
the only special thing about them is that they are complete as normed spaces.

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.
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?

matt grime