Show Self-Adjointness of Operator E on $L_{2}$ of Square Integrable Functions

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In summary, we need to show that the operator E defined by, for f in $L_{2}$, $(Ef)(x)=0.5(f(x)+f(-x)$ is self-adjoint. This is equivalent to showing that $\langle Ef | g \rangle = \langle f | Eg \rangle$ for all $f,g\in L_{2}$. To do this, we can use a Riemann integral and the fact that square integrable functions vanish at infinity. By letting $y=-x$, we can see that the two integrals are equal, thus proving that E is self-adjoint.
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Fermat1
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Conisider the space $L_{2}$ of square integrable functions on R with the usual integral inner product. Show that the operator E defined by, for f in $L_{2}$,
$(Ef)(x)=0.5(f(x)+f(-x)$ is self adoint.

It seems that in order for this to be true we have that f(-x) is the conjugate of f(x) but I don't know why this is true.
 
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  • #2
So, we need to show that $\langle Ef | g \rangle = \langle f | Eg \rangle$ for all $f,g\in L_{2}$. This is tantamount to showing that
$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu
= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$
or
$$ \int_{ \mathbb{R}}\left[ f^{*}(x) g(x)+f^{*}(-x) g(x) \right]\, d\mu
= \int_{\mathbb{R}} \left[f^{*}(x) g(x)+f^{*}(x)g(-x) \right] \, d\mu,$$
or
$$\int_{ \mathbb{R}} f^{*}(-x) g(x)\, d\mu
= \int_{\mathbb{R}} f^{*}(x)g(-x) \, d\mu.$$
Can you think of a way to show this?
 
  • #3
Ackbach said:
So, we need to show that $\langle Ef | g \rangle = \langle f | Eg \rangle$ for all $f,g\in L_{2}$. This is tantamount to showing that
$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu
= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$
or
$$ \int_{ \mathbb{R}}\left[ f^{*}(x) g(x)+f^{*}(-x) g(x) \right]\, d\mu
= \int_{\mathbb{R}} \left[f^{*}(x) g(x)+f^{*}(x)g(-x) \right] \, d\mu,$$
or
$$\int_{ \mathbb{R}} f^{*}(-x) g(x)\, d\mu
= \int_{\mathbb{R}} f^{*}(x)g(-x) \, d\mu.$$
Can you think of a way to show this?

I had thought of using parts and then using the fact that square integrable functions vanish at + and - infinity but sine there are no derivatives involved that looks to be a non-starter.
 
  • #4
Fermat said:
I had thought of using parts and then using the fact that square integrable functions vanish at + and - infinity but sine there are no derivatives involved that looks to be a non-starter.

Well, think of the integral (temporarily) as a Riemann integral:
$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$
What happens when you let $y=-x$?
 
  • #5
Ackbach said:
Well, think of the integral (temporarily) as a Riemann integral:
$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$
What happens when you let $y=-x$?

you get the equality.
 

FAQ: Show Self-Adjointness of Operator E on $L_{2}$ of Square Integrable Functions

What is an operator in the context of self-adjointness?

An operator in mathematics is a function that maps one mathematical object to another. In the context of self-adjointness, an operator is a function that maps a vector space to itself, and can be thought of as a transformation on that vector space.

What is the significance of self-adjointness in linear algebra?

Self-adjointness is an important property of operators in linear algebra because it ensures that the operator is symmetric with respect to the inner product of the vector space. This means that the operator and its adjoint (conjugate transpose) are equal, and therefore, the operator can be easily studied and analyzed using linear algebra techniques.

How is self-adjointness related to Hermitian operators?

In the context of linear algebra, a self-adjoint operator is equivalent to a Hermitian operator. This means that the operator is equal to its adjoint, which is the complex conjugate of the original operator. Hermitian operators have many useful properties, including that they have real eigenvalues and orthogonal eigenvectors.

How is self-adjointness shown for an operator on $L_{2}$ of square integrable functions?

To show self-adjointness for an operator on $L_{2}$ of square integrable functions, we must demonstrate that the operator is equal to its adjoint (conjugate transpose) with respect to the inner product of the function space. This can be done by manipulating the integral representation of the operator and using properties of the inner product and integration.

Why is self-adjointness important in quantum mechanics?

In quantum mechanics, self-adjoint operators play a crucial role as observables. This means that the eigenvalues of the operator correspond to measurable quantities in the physical system. Self-adjoint operators also have real eigenvalues, which are necessary for the probabilistic interpretation of quantum mechanics.

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