Is a Self-adjoint Operator Always Real?

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In summary, the conversation discusses the properties of a linear operator and its adjoint, focusing on the relationship between the scalar product and the self-adjointness of L. The conclusion is that <f,Lf> is a real value, as shown by the equation being equal to its complex conjugate. This is a common result seen in quantum mechanics with self-adjoint operators representing real valued observables.
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daudaudaudau
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Hello.

I have a linear operator, [tex]L[/tex], and its adjoint [tex]L^a[/tex]. [tex]L[/tex] is self-adjoint, so [tex]L=L^a[/tex]. I'm being told that the following is true:

[tex]\langle f,Lh\rangle=\langle Lf,h\rangle[/tex].

But what if the scalar product is not the symmetric product? What if

[tex]\langle f,h\rangle=\langle h,f\rangle^*[/tex]

where [tex]^*[/tex] is complex conjugation ? Then my first equation tells me that

[tex]\langle f,Lf\rangle=\langle Lf,f\rangle[/tex].

and the second one says that

[tex]\langle f,Lf\rangle=\langle Lf,f\rangle^*[/tex].

But which is true?
 
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  • #2
daudaudaudau said:
But which is true?

Both, <f,Lf> is real.
 
  • #3
But how do you know?
 
  • #4
daudaudaudau said:
But how do you know?

You just showed that it equals its complex conjugate, <f,Lf> = <Lf,f>* = <f,Lf>*. So it is real.
Quite a standard result (eg, self adjoint operators represent real valued observables in quantum mechanics).
 

What is a self-adjoint operator?

A self-adjoint operator is a type of linear operator in mathematics that has the property of being equal to its own adjoint. This means that the operator and its adjoint have the same matrix representation, and thus have the same eigenvalues and eigenvectors. In simpler terms, a self-adjoint operator is symmetric with respect to a given inner product.

What is the difference between a self-adjoint operator and a Hermitian operator?

A self-adjoint operator and a Hermitian operator are often used interchangeably, but there is a subtle difference between the two. A self-adjoint operator is defined on a real vector space, while a Hermitian operator is defined on a complex vector space. Additionally, a self-adjoint operator satisfies the stronger condition of being equal to its own adjoint, while a Hermitian operator only satisfies the condition of being equal to its conjugate transpose.

Why are self-adjoint operators important in quantum mechanics?

In quantum mechanics, self-adjoint operators are used to represent physical observables such as position, momentum, and energy. This is because the eigenvalues and eigenvectors of a self-adjoint operator correspond to the possible outcomes and states of a quantum system. Additionally, the spectral theorem states that any self-adjoint operator on a complex Hilbert space has a complete set of eigenvectors, making it a useful tool in solving quantum mechanical problems.

Can a non-self-adjoint operator be made self-adjoint?

No, a non-self-adjoint operator cannot be made self-adjoint. This is because the self-adjoint property is a fundamental property of an operator and cannot be changed. However, a non-self-adjoint operator can be transformed into a similar self-adjoint operator through a process called diagonalization, which involves finding a basis of eigenvectors for the operator.

What is the physical significance of the self-adjointness condition?

The self-adjointness condition is important in quantum mechanics because it ensures that the operator representing a physical observable is Hermitian, or self-adjoint. This guarantees that the eigenvalues of the operator will be real, which corresponds to measurable physical quantities. The self-adjointness condition also ensures that the operator has a complete set of eigenvectors, allowing for the spectral decomposition of the operator and simplifying calculations in quantum mechanics.

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