Understanding the Definition of the Adjoint Operator in Linear Algebra

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In summary, there is a difference between self-adjoint operators and Hermitian operators. Mathematicians tend to talk more abstractly about self-adjoint operators while physicists talk mostly about Hermitian operators. An operator is Hermitian if it is self-adjoint on a vector space over the complex field. The definition of the adjoint may vary depending on whether the operator is bounded or not, but in general it is defined as a linear transformation between dual spaces.
  • #1
jostpuur
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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?
 
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  • #2
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
HallsofIvy said:
In particular if U= V and T*= T, that is, if <Tu,v>= <u,Tv>, then T is "self-adjoint".

And it is allowable to call T Hermitian only if U is a vector space over complex field?
 
  • #4
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
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.
 
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  • #6
Galileo said:
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.

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.
 
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  • #7
jostpuur said:
And it is allowable to call T Hermitian only if U is a vector space over complex field?

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.
 
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  • #8
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

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

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

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

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

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.

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->H and T*:H->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 :redface:
 
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What is the difference between self-adjoint and hermitian operators?

Self-adjoint and hermitian operators are very similar, but there is a subtle difference between them. A self-adjoint operator is one that is equal to its own adjoint, meaning that the operator and its adjoint have the same eigenvalues. On the other hand, a hermitian operator is a complex linear operator that is equal to its own conjugate transpose, meaning that the operator and its conjugate transpose have the same eigenvalues. In simpler terms, a self-adjoint operator is real and a hermitian operator is complex.

Can a self-adjoint operator be non-hermitian?

No, a self-adjoint operator must also be hermitian. This is because a self-adjoint operator is defined as one that is equal to its own adjoint, and the adjoint of a hermitian operator is its conjugate transpose. Therefore, a self-adjoint operator must also be complex and have the same eigenvalues as its conjugate transpose, making it hermitian.

What are the properties of self-adjoint and hermitian operators?

There are several important properties of self-adjoint and hermitian operators. These include:
1. They have real eigenvalues
2. They have orthogonal eigenvectors
3. They are diagonalizable
4. They are normal operators
5. They have a complete set of eigenfunctions
6. They have a real spectrum
7. Their eigenvalues are non-degenerate (except for degenerate operators)
These properties make self-adjoint and hermitian operators useful in quantum mechanics, as they simplify calculations and have physical interpretations.

How are self-adjoint and hermitian operators related to quantum mechanics?

In quantum mechanics, self-adjoint and hermitian operators are used to represent observable quantities, such as position, momentum, and energy. This is because their eigenvalues are real and correspond to the possible outcomes of measurements. Additionally, the eigenvectors of these operators represent the states of the system, and the coefficients of the eigenvectors in the wave function represent the probability amplitudes of the system being in that state.

What is the significance of the self-adjoint and hermitian properties in linear algebra?

The self-adjoint and hermitian properties are important in linear algebra because they guarantee that the operator has a complete set of orthogonal eigenvectors, making it easy to diagonalize. This allows for simpler calculations and interpretations, particularly in quantum mechanics. Additionally, these properties ensure that the operator is normal, meaning it commutes with its adjoint, which is useful in many applications.

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