Is the Adjoint of an Operator the Same as the Original Operator?

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In summary: The problem is not easy,if it were,we'd not be doing mathematics here,but something totally different.I'm not familiar with nuclear subspaces, but I tried my method and it seemed to work, if you define the adjoint here as the operator such that: \int dx f^{*}(x) \hat L ( g(x) ) = \int dx (\hat L^{\dagger} (f(x)))^{*} g(x)
  • #1
Feynmanfan
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Could you help me calculate the adjoint of this operator?: Â= x d/dx

I don't know where to start from. I am asked to say if it can be an observable. Well if I find that the adjoint is the same then the operator is Hermitian, right? It can only be observable if it's hermitian.

Thanks!
 
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  • #2
What's the inner product? without knowign that the question is impossible.

It is the operator L such that <Lf,g> = <f,xg'> by definition, so just do it ie work it out (i'm guessing it's some L^2 kind of space).
 
  • #3
I can assume you're dealing with a rigged Hilbert space in which the nuclear subspace is the the set of tempered distributions of [itex] \mathbb{R} [/itex].

The basis in the Hilbert space is made up of Hermite polynomials and the scalar product is defined

[tex](\phi,\psi)=:\int \bar{\phi}(x) \psi (x) \ dx [/tex].

Here are 3 nice articles reccomended by mr.Bogolubov

Kristensen,P.,Mejlbo L.,Poulsen,E.T. (1964-1967)

"Tempered Distributions in Infinitely Many Dimensions:I Canonical Field Operators",Commun.Math.Phys.,1,175.

"II.Displacement Operators",Math.Scand.,14,129.

"III.Linear Transformations of Field Operators",Commun.Math.Phys.,6,29.


Daniel.
 
  • #4
As for being an observable or not,it must be
1.linear.
2.densly-defined.
3.self-adjoint.

Daniel.
 
  • #5
The operator can be written as [itex]\frac{i}{\hbar}\hat x \hat p[/itex], so try taking the adjoint of that (you can ignore h if you want, or set it to 1). Remember x and p are self adjoint, take the complex conjugate of the constant out front, and, just like with transposed matrices, the order of the operators needs to be reversed, ie, (Adj(AB)=Adj(B)Adj(A)).
 
  • #6
That's not true in this case.

[tex] \hat{p}^{\dagger}\hat{x}^{\dagger}\subset\left(\hat{x}\hat{p}\right)^{\dagger} [/tex]

,because both operators are unbounded on the Hilbert space of states (spanned by Hermite polynomials,check post #3) and therefore can be defined only on an everywhere dense subset of [itex] \mathcal{H} [/itex],called nuclear subspace.

Things are not really easy.:wink:

Daniel.
 
  • #7
I'm not familiar with nuclear subspaces, but I tried my method and it seemed to work, if you define the adjoint here as the operator such that:

[tex] \int dx f^{*}(x) \hat L ( g(x) ) = \int dx (\hat L^{\dagger} (f(x)))^{*} g(x) [/tex]

where the integral is taken over the possible range of the particle (usually -[itex]\infty[/itex] to +[itex]\infty[/itex]), and the functions are assumed to vanish at the boundaries. I doubt if the question intended for anything more general than this.
 
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  • #8
The question asked to compute the adjoint.I think it's pretty clear.One has to apply the definition of the adjoint,to see whether the adjoint exists and under what circumstances.

The problem is not easy,if it were,we'd not be doing mathematics here,but something totally different.

Daniel.
 
  • #9
StatusX said:
I'm not familiar with nuclear subspaces, but I tried my method and it seemed to work, if you define the adjoint here as the operator such that:

[tex] \int dx f^{*}(x) \hat L ( g(x) ) = \int dx (\hat L^{\dagger} (f(x)))^{*} g(x) [/tex]

where the integral is taken over the possible range of the particle (usually -[itex]\infty[/itex] to +[itex]\infty[/itex]), and the functions are assumed to vanish at the boundaries. I doubt if the question intended for anything more general than this.
How would you do this? It would seem that we need to have:

[tex]f^{*}(x) \hat L ( g(x) ) = (\hat L^{\dagger} (f(x)))^{*} g(x) [/tex]

[tex]xf^{*}(x) g'(x) = (\hat L^{\dagger} (f(x)))^{*} g(x) [/tex]

[tex]\left (\frac{xf^{*}(x) g'(x)}{g(x)}\right )^{*} = \hat L^{\dagger} (f(x))[/tex]

[tex]\left (\frac{If^{*}g'}{g}\right )^{*} = \hat L^{\dagger} (f)[/tex]

where I is the identity mapping. This doesn't seem to be well-defined, since [itex]L^{\dagger} (f)[/itex] depends on g, so I assume this is not the right way to find [itex]L^{\dagger}[/itex]. So how do you do it?
 
  • #10
I used my suggestion above (using the x and p operators) to derive the adjoint, and then I verified that it satisfied that condition. In general, the integrands are not equal, and in fact my idea only works if you assume the function vanishes at the endpoints.
 
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  • #11
i think, daniel, that you're making it far more complicated than it should be given it was homework.

it ought to be quite straight forward, if we had full information, such as what the space is and the inner product.
 
  • #12
Maybe so,but i made my point.

Daniel.
 

What is the adjoint of an operator?

The adjoint of an operator is a mathematical concept that applies to linear operators on vector spaces. It is defined as the transpose of the operator's matrix representation with respect to a chosen basis.

Why is the adjoint of an operator important?

The adjoint of an operator is important because it allows us to define and study important properties of linear operators, such as self-adjointness, orthogonality, and eigenvalues. It also plays a key role in applications of linear algebra, such as quantum mechanics and signal processing.

How is the adjoint of an operator calculated?

The calculation of the adjoint of an operator involves finding the transpose of its matrix representation with respect to a chosen basis. This can be done by taking the complex conjugate of the matrix elements and then swapping the rows and columns of the matrix.

What is the relationship between the adjoint of an operator and its inverse?

For a linear operator, the adjoint and the inverse are closely related. If an operator is invertible, then its inverse is equal to its adjoint divided by the determinant of the operator. This relationship is known as the adjoint-inverse theorem.

What is the difference between the adjoint of an operator and the transpose of a matrix?

The adjoint of an operator is a more general concept than the transpose of a matrix. While the transpose involves only the swapping of rows and columns of a matrix, the adjoint also involves taking the complex conjugate of the matrix elements. Additionally, the adjoint can be defined for any linear operator, while the transpose is only defined for matrices.

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