Is the Operator Hermitian?

[Lambda96]

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

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.

 

[PeroK]

Note that $\frac{d^2}{dx^2} =\frac{d}{dx}\frac{d}{dx}$

 

[vela]

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.

 

[Lambda96]

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

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

 

[PeroK]

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.

 

[PeroK]

Actually, if you just process the first term, things should cancel out. And things fall out in 5-6 lines.

 

[PeroK]

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!

 

[Lambda96]

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

 

[PeroK]

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:
$$-\bigl< xf, \frac{d}{dx}g \bigr>+\bigl< \frac{d}{dx}(xf), \frac{d}{dx}g \bigr>$$

 

[vela]

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

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.

 

[Lambda96]

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.

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

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

 

[Lambda96]

Thanks PeroK and vela for your help and for looking over my calculation