Is the Operator Hermitian?

In summary, the conversation discusses a math problem involving properties and identities. The problem involves finding the value of <Lf, g> using the given identities. The conversation talks about rewriting terms using the identities and applying specific rules in order to obtain the final solution. However, there are difficulties in applying the rules correctly and obtaining the desired result.
  • #1
Lambda96
189
65
Homework Statement
##\bigl\langle f,\cal L g \bigr\rangle=\bigl\langle \cal L f,\ g \bigr\rangle##
Relevant Equations
none
Hi,

unfortunately, I have problems with the following task

Bildschirmfoto 2023-06-22 um 12.37.30.png

I tried the fast way, unfortunately I have problems with it

I have already proved the following properties, ##\bigl< f,xg \bigr>=\bigl< xf,g \bigr>## and ##\bigl< f, \frac{d}{dx}g \bigr>=-\overline{f(0)} g(0)+\bigl< f,g \bigr>-\bigl< \frac{d}{dx}f,g \bigr>## and then proceeded as follows:

$$\bigl< f,\cal L g \bigr>$$
$$\bigl< f,(-x\frac{d^2}{dx^2}+(x-1)\frac{d}{dx})g \bigr>$$
$$\bigl< f,-x \frac{d^2}{dx^2}g+x \frac{d}{dx}g-\frac{d}{dx}g \bigr>$$
$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< f,x \frac{d}{dx}g \bigr>-\bigl< f,\frac{d}{dx}g \bigr>$$

Then for the last term, I used the above identity ##\bigl< f,\frac{d}{dx}g \bigr>=-\overline{f(0)} g(0)+\bigl< f,g \bigr>-\bigl< \frac{d}{dx}f,g \bigr>## and obtained the following:

$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< f,x \frac{d}{dx}g \bigr>+\overline{f(0)} g(0)- \bigl< f,g \bigr>+\bigl< \frac{d}{dx}f,g \bigr>$$

Unfortunately now I'm stuck because I don't know what to do with the first two terms, in the hint it says to apply the above identities to ##\bigl< f,-x \frac{d^2}{dx^2}g \bigr>## but unfortunately I don't know how.
 
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  • #2
Note that ##\frac{d^2}{dx^2} =\frac{d}{dx}\frac{d}{dx}##
 
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  • #3
For the second term, use the first property to move the ##x## onto ##f##. That is, you can say that
$$\langle f, x\frac{d}{dx}g \rangle = \langle xf, \frac{d}{dx}g\rangle = \langle h, \frac{d}{dx}g\rangle,$$ where ##h(x) = x f(x)##. Use the second property to move the ##d/dx## over onto ##h##. And so on.
 
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  • #4
Thanks PeroK and vela for your help 👍👍

I tried to apply your tips and got the following:

$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< f,x \frac{d}{dx}g \bigr>+\overline{f(0)} g(0)- \bigl< f,g \bigr>+\bigl< \frac{d}{dx}f,g \bigr>$$.
$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< xf, \frac{d}{dx}g \bigr>+\overline{f(0)} g(0)- \bigl< f,g \bigr>+\bigl< \frac{d}{dx}f,g \bigr>$$
$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< h, \frac{d}{dx}g \bigr>+\overline{f(0)} g(0)- \bigl< f,g \bigr>+\bigl< \frac{d}{dx}f,g \bigr>$$
$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>+\bigl< h,g \bigr>-\bigl< \frac{d}{dx}h,g \bigr>+\overline{f(0)} g(0)- \bigl< f,g \bigr>+\bigl< \frac{d}{dx}f,g \bigr>$$

I then tried to rewrite the term ##\bigl< f,-x \frac{d^2}{dx^2}g \bigr>## as follows ##-\bigl< h, \frac{d}{dx} \frac{d}{dx}g \bigr>## but I am not sure if I can now apply the second identity from the hint to it and get the following

$$-\bigl< h, \frac{d}{dx} \frac{d}{dx}g \bigr>=-\bigl< \frac{d}{dx}h,g \bigr>+\bigl< \frac{d^2}{dx^2}h,g \bigr>$$
 
  • #5
First, it's a good idea to write down what you are aiming for:
$$\langle L f,\ g \rangle =- \langle (x\frac{d^2}{dx^2}) f, g \rangle + \langle (x\frac{d}{dx}) f, g \rangle - \langle \frac{d}{dx} f, g \rangle$$Now$$\langle f,\ Lg \rangle =- \langle f, \ (x\frac{d^2}{dx^2})g \rangle + \langle f, \ (x\frac{d}{dx})g \rangle - \langle f, \ \frac{d}{dx}g \rangle$$You need to apply the "##x##" rule to the first and second terms and the ##\frac d {dx}## rule to the third term. Then apply the ##\frac d {dx}## rule twice to the new first term and once to the middle term. And see what you get.
 
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  • #6
Actually, if you just process the first term, things should cancel out. And things fall out in 5-6 lines.
 
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  • #7
vela said:
For the second term, use the first property to move the ##x## onto ##f##. That is, you can say that
$$\langle f, x\frac{d}{dx}g \rangle = \langle xf, \frac{d}{dx}g\rangle = \langle h, \frac{d}{dx}g\rangle,$$ where ##h(x) = x f(x)##. Use the second property to move the ##d/dx## over onto ##h##. And so on.
Actually, it turns out to be simpler to leave the second and third terms alone and watch them vanish!
 
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  • #8
Thanks PeroK for your help 👍, I have now tried to rewrite only the first term using the two identities in the hint, unfortunately I have some difficulties with this

$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>$$
$$-\bigl< xf, \frac{d^2}{dx^2}g \bigr>$$
$$-\bigl< xf, \frac{d}{dx}\frac{d}{dx}g \bigr>$$
$$-\bigl< xf, \frac{d}{dx}g \bigr>+\bigl< x\frac{d}{dx}f, \frac{d}{dx}g \bigr>$$
$$-\bigl< xf, g \bigr>+\bigl< x\frac{d}{dx}f, g \bigr>+\bigl< x\frac{d}{dx}f, g \bigr>-\bigl< x\frac{d^2}{dx^2}f, g \bigr>$$

If I now combine this expression with the other two terms, i.e. ##\bigl< f,x \frac{d}{dx}g \bigr>## and ##\bigl< f, \frac{d}{dx}g \bigr>##, I unfortunately do not get the final result, which is why I assume that I have applied the rule ##\frac{d}{dx}## twice incorrectly on the first term
 
  • #9
Lambda96 said:
Thanks PeroK for your help 👍, I have now tried to rewrite only the first term using the two identities in the hint, unfortunately I have some difficulties with this

$$\bigl< f,-x \frac{d^2}{dx^2}g \bigr>$$
$$-\bigl< xf, \frac{d^2}{dx^2}g \bigr>$$
$$-\bigl< xf, \frac{d}{dx}\frac{d}{dx}g \bigr>$$
$$-\bigl< xf, \frac{d}{dx}g \bigr>+\bigl< x\frac{d}{dx}f, \frac{d}{dx}g \bigr>$$
This is not right. It should be:
$$-\bigl< xf, \frac{d}{dx}g \bigr>+\bigl< \frac{d}{dx}(xf), \frac{d}{dx}g \bigr>$$
 
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  • #10
Lambda96 said:
$$-\bigl< xf, \frac{d}{dx}g \bigr>+\bigl< x\frac{d}{dx}f, \frac{d}{dx}g \bigr>$$
The reason I introduced the function ##h=xf## earlier was in hopes of you avoiding the mistake you made. In terms of ##h##, you should get
$$-\langle xf, \frac{d}{dx}g' \rangle = -\langle h, \frac{d}{dx}g' \rangle = \overline{h(0)}g'(0) - \langle h,g' \rangle + \langle \frac{d}{dx}h,g'\rangle.$$ Then if you substitute back in ##xf## for ##h##, you'll end up with what @PeroK did in the previous post.
 
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  • #11
Thanks PeroK and vela for your help 👍👍

I have now continued the notation with ##h=xf## and got the following.

$$-\bigl< h, \frac{d^2}{dx^2}g \bigr>=-\bigl< h, \frac{d}{dx}g \bigr>+\bigl< \frac{d}{dx}h, \frac{d}{dx}g \bigr>$$

I then substituted ##h=xf## in the second term and applied the product rule

$$=-\bigl< h, \frac{d}{dx}g \bigr>+\bigl< x\frac{d}{dx}f, \frac{d}{dx}g \bigr>+\bigl< f, \frac{d}{dx}g \bigr>$$

I then applied the ##\frac{d}{dx}## rule from the hint again for the second term

$$=-\bigl< h, \frac{d}{dx}g \bigr>+\bigl<x\frac{d}{dx}f,g \bigr>-\bigl<\frac{d}{dx} \bigl(x \frac{d}{dx}f\bigr),g \bigr>+\bigl<f, \frac{d}{dx}g \bigr>$$.

Then applied the product rule for the third term

$$=-\bigl< h, \frac{d}{dx}g \bigr>+\bigl<x\frac{d}{dx}f,g \bigr>-\bigl\langle x \frac{d^2}{dx^2}f,g \bigr\rangle -\bigl\langle \frac{d}{dx} f ,g \bigr\rangle +\bigl< f, \frac{d}{dx}g \bigr>$$

Then I applied ##h=xf## for the first term and applied the x rule from the hint again

$$=-\bigl< f, x\frac{d}{dx}g \bigr>+\bigl<x\frac{d}{dx}f,g \bigr>-\bigl\langle x \frac{d^2}{dx^2}f,g \bigr\rangle -\bigl\langle \frac{d}{dx} f ,g \bigr\rangle +\bigl< f, \frac{d}{dx}g \bigr>$$

If I now add the two unchanged terms to this, I get the following.

$$=-\bigl< f, x\frac{d}{dx}g \bigr>+\bigl<x\frac{d}{dx}f,g \bigr>-\bigl\langle x \frac{d^2}{dx^2}f,g \bigr\rangle -\bigl\langle \frac{d}{dx} f , g \bigr\rangle +\bigl< f, \frac{d}{dx}g \bigr>+\bigl< f,x \frac{d}{dx}g \bigr>-\bigl< f,\frac{d}{dx}g \bigr>$$

$$-\bigl\langle x \frac{d^2}{dx^2}f,g \bigr\rangle+\bigl<x\frac{d}{dx}f,g \bigr>-\bigl\langle \frac{d}{dx} f ,g \bigr\rangle=\bigl\langle \cal L f,x \bigr\rangle$$
 
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  • #12
Thanks PeroK and vela for your help and for looking over my calculation 👍👍
 

FAQ: Is the Operator Hermitian?

What is a Hermitian operator?

A Hermitian operator, also known as a self-adjoint operator, is an operator in quantum mechanics that is equal to its own adjoint. Mathematically, an operator \( \hat{A} \) is Hermitian if \( \hat{A} = \hat{A}^\dagger \), where \( \hat{A}^\dagger \) is the conjugate transpose of \( \hat{A} \). Hermitian operators have real eigenvalues and their eigenvectors form a complete orthonormal basis.

How do you show that an operator is Hermitian?

To show that an operator \( \hat{A} \) is Hermitian, you need to demonstrate that \( \langle \psi | \hat{A} \phi \rangle = \langle \hat{A} \psi | \phi \rangle \) for all vectors \( | \psi \rangle \) and \( | \phi \rangle \) in the Hilbert space. This means that the inner product of \( \hat{A} \phi \) with \( \psi \) must equal the inner product of \( \phi \) with \( \hat{A} \psi \). If this condition holds for all possible \( | \psi \rangle \) and \( | \phi \rangle \), then \( \hat{A} \) is Hermitian.

Why are Hermitian operators important in quantum mechanics?

Hermitian operators are crucial in quantum mechanics because they correspond to observable physical quantities. The eigenvalues of Hermitian operators are real, which is necessary because measurements of physical quantities must yield real numbers. Additionally, the eigenvectors of Hermitian operators form an orthonormal basis, which ensures that the quantum states can be described consistently and completely.

Can you give an example of a Hermitian operator?

An example of a Hermitian operator is the Hamiltonian operator \( \hat{H} \) in quantum mechanics, which represents the total energy of the system. Another common example is the position operator \( \hat{x} \) and the momentum operator \( \hat{p} \), both of which are Hermitian. For instance, the momentum operator in one dimension is given by \( \hat{p} = -i\hbar \frac{d}{dx} \), and it can be shown to be Hermitian by integration by parts, assuming appropriate boundary conditions.

What are the properties of Hermitian operators?

Hermitian operators have several important properties:1. Their eigenvalues are real.2. Eigenvectors corresponding to different eigenvalues are orthogonal.3. They can be diagonalized by a unitary transformation

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