Determining whether an operator is Hermitian

In summary: I was just trying to use something from a previous proof but this is so much easier. I got it now. Thanks!In summary, the conversation discusses determining which linear operators are Hermitian for a set of functions defined on the interval -∞< x < ∞ that go to zero faster than 1/x for x → ±∞. The homework also includes determining the eigenvalues for these operators. The solution involves using an inner product and checking if the operators satisfy the Hermitian property.
  • #1
GoKush
2
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Homework Statement



Consider the set of functions [itex] {f(x)} [/itex] of the real variable [itex] x [/itex] de fined on the interval [itex] -\infty< x < \infty [/itex] that go
to zero faster than [itex] 1/x [/itex] for [itex]x\rightarrow ±\infty[/itex] , i.e.,
[tex]
\lim_{n\rightarrow ±\infty} {xf(x)}=0
[/tex]
For unit weight function, determine which of the following linear operators is Hermitian when acting upon [itex] {f(x)} [/itex]:
[tex] (a) \frac{d}{dx} + x[/tex] [tex](b) -i \frac{d}{dx}+x^2 [/tex][tex](c) ix \frac{d}{dx} [/tex][tex](d) ix \frac{d^3}{dx^3} .[/tex]


Homework Equations



[itex] Hf(x)=λf(x) [/itex] has real values of [itex] λ [/itex] where [itex] H [/itex] is a Hermitian operator and [itex] λ [/itex] are it's eigenvalues



The Attempt at a Solution



[tex] a) \frac{df(x)}{dx} + xf(x) = λf(x) [/tex][tex] \frac{df(x)}{dx} + (x-λ)f(x) = 0 [/tex] [tex]\text{Integrating factor is }e^{\int (x-λ)dx}=e^{\frac{1}{2} x^2-λx}[/tex]
[tex]e^{\frac{1}{2} x^2-λx}f(x)=constant[/tex]

I've done a similar thing for parts a), b), and c) but I'm not sure what to do with this or if it even helps. For d) I've tried to work out the eigenfunctions but get to mess and didn't really want to continue down the route I was going without knowing if this was useful or not.

[tex]\text{b) leads to } e^{ix(\frac{x^2}{3}-λ)}f(x)=constant[/tex]
[tex]\text{c) leads to } x^{iλ}f(x)=constant[/tex]
[tex]\text{d) } ix \frac{d^3f(x)}{dx^3}=λf(x) [/tex]
[tex] x \frac{d^3f(x)}{dx^3}+iλf(x)=0 [/tex]
[tex]\text{let }x=f(t)[/tex]
[tex]\frac{d^3f(x)}{dx^3}=\frac{d^3f(x)}{dt^3}\frac{d^3t}{dx^3} [/tex]
[tex]\text{want }\frac{d^3t}{dx^3}=\frac{1}{x}[/tex]
[tex]\text{(After integrating 3 times }t=\frac{1}{2}(x^2(ln(x)-\frac{3}{2})) [/tex]

This is were I decided not to continue until I knew whether or not I was actually doing anything right.

Thanks in advance for any help.
 
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  • #2
GoKush said:

Homework Statement



Consider the set of functions [itex] {f(x)} [/itex] of the real variable [itex] x [/itex] de fined on the interval [itex] -\infty< x < \infty [/itex] that go
to zero faster than [itex] 1/x [/itex] for [itex]x\rightarrow ±\infty[/itex] , i.e.,
[tex]
\lim_{n\rightarrow ±\infty} {xf(x)}=0
[/tex]
For unit weight function, determine which of the following linear operators is Hermitian when acting upon [itex] {f(x)} [/itex]:
[tex] (a) \frac{d}{dx} + x[/tex] [tex](b) -i \frac{d}{dx}+x^2 [/tex][tex](c) ix \frac{d}{dx} [/tex][tex](d) ix \frac{d^3}{dx^3} .[/tex]


Homework Equations



[itex] Hf(x)=λf(x) [/itex] has real values of [itex] λ [/itex] where [itex] H [/itex] is a Hermitian operator and [itex] λ [/itex] are it's eigenvalues



The Attempt at a Solution



[tex] a) \frac{df(x)}{dx} + xf(x) = λf(x) [/tex][tex] \frac{df(x)}{dx} + (x-λ)f(x) = 0 [/tex] [tex]\text{Integrating factor is }e^{\int (x-λ)dx}=e^{\frac{1}{2} x^2-λx}[/tex]
[tex]e^{\frac{1}{2} x^2-λx}f(x)=constant[/tex]

I've done a similar thing for parts a), b), and c) but I'm not sure what to do with this or if it even helps. For d) I've tried to work out the eigenfunctions but get to mess and didn't really want to continue down the route I was going without knowing if this was useful or not.

[tex]\text{b) leads to } e^{ix(\frac{x^2}{3}-λ)}f(x)=constant[/tex]
[tex]\text{c) leads to } x^{iλ}f(x)=constant[/tex]
[tex]\text{d) } ix \frac{d^3f(x)}{dx^3}=λf(x) [/tex]
[tex] x \frac{d^3f(x)}{dx^3}+iλf(x)=0 [/tex]
[tex]\text{let }x=f(t)[/tex]
[tex]\frac{d^3f(x)}{dx^3}=\frac{d^3f(x)}{dt^3}\frac{d^3t}{dx^3} [/tex]
[tex]\text{want }\frac{d^3t}{dx^3}=\frac{1}{x}[/tex]
[tex]\text{(After integrating 3 times }t=\frac{1}{2}(x^2(ln(x)-\frac{3}{2})) [/tex]

This is were I decided not to continue until I knew whether or not I was actually doing anything right.

Thanks in advance for any help.

No, that's really not the right way to go about it. You've likely defined an inner product ##<f,g>= \int_{-\infty}^\infty f^*(x) g(x) dx ##. In terms of that inner product H is Hermitian if <f,Hg>=<Hf,g> for any two functions f and g. Does that sound familiar? Checking that is the way to check if an operator is Hermitian. Start with the two parts of the first one. Is x Hermitian? That's pretty easy. Now is d/dx Hermitian? That's a little harder. Try looking at an integration by parts.
 
  • #3
Oh, thank you I knew I must've been doing something wrong.
 

FAQ: Determining whether an operator is Hermitian

1. What is a Hermitian operator?

A Hermitian operator is a linear operator on a complex vector space that is equal to its own conjugate transpose. In other words, the operator is self-adjoint, meaning it is equal to its own adjoint.

2. How can you determine if an operator is Hermitian?

To determine if an operator is Hermitian, you can check if it is equal to its own conjugate transpose. This can be done by taking the complex conjugate of the matrix representing the operator and comparing it to the transpose of the matrix. If they are equal, then the operator is Hermitian.

3. What is the significance of a Hermitian operator?

Hermitian operators have many important properties in quantum mechanics, such as having real eigenvalues and being associated with physical observables. In addition, they can be used to describe the time evolution of a quantum system.

4. Can a non-square matrix be Hermitian?

No, a non-square matrix cannot be Hermitian. Hermitian operators must have the same number of rows and columns, as they are defined as being equal to their own conjugate transpose, which requires the matrix to be square.

5. Are all physical observables represented by Hermitian operators?

Yes, in quantum mechanics, all physical observables must be represented by Hermitian operators. This is because the eigenvalues of a Hermitian operator are real and correspond to the possible outcomes of a measurement, making them suitable for representing physical quantities.

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