Where can I learn how to manipulate operators?

  • Thread starter Thread starter k4ff3
  • Start date Start date
  • Tags Tags
    Operators
k4ff3
Messages
39
Reaction score
0
Nice to be back here at PF and to physics after a year off in the software industry. Now it's time to catch up again :)

I feel like I never really get the grasp of manipulating operators. In QM there's a lot of trixing and mixing going on, and I really would like to learn to do the magic. Oddly, none of my books deal with these techniques, they just assume that you know them.

I can give a simple example of a wave equation:

\frac{ \partial^2 \eta }{\partial x^2} - \frac{ \partial^2 \eta }{c \cdot \partial t^2} = ( \frac{ \partial}{\partial x} - \frac{ \partial}{c \cdot \partial x} )( \frac{ \partial}{\partial x} + \frac{ \partial}{c \cdot \partial x} ) \eta = 0 (*)

By letting u=x-ct and v=x+ct one can show that

\frac{ \partial}{\partial x} - \frac{ \partial}{c \cdot \partial t} = 2 \frac{ \partial}{ \partial u} (**)

\frac{ \partial}{\partial x} + \frac{ \partial}{c \cdot \partial t} = 2 \frac{ \partial}{ \partial v} (***)

by the difference of two squares.

( Which simplifies (*). )

But I don't get how the operator formulas (**) and (***) are obtained! It feels terrible to miss out on something this basic.

Any suggestions?
 
Last edited:
Physics news on Phys.org
Corrected. Thanks.

Still stuck though. Despite the article. I know how to use the chain rule when I have variables. But it's when I'm dealing purely with operators I struggle. I.e

\frac{ \partial}{\partial u}

How can you apply a chain rule on this guy?
 
Consider u and v as functions of x and t: u(x,t) and v(x,t).
Then apply the chain rule as is done in the example in the section I've linked.
 
So for f a function of u and v, which are themselves a function of x and t you have according to the chain rule:

\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}

Also:

\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial t}=c\frac{\partial f}{\partial v}-c\frac{\partial f}{\partial u}

Now you can multiply the second equation by 1/c and subtract it from the first to get **, and add to get ***.

You put the trial function f in there to see what happens, and then remove it at the end.
 
Exactly!
 
Be a quantum mechanic.
 
k4ff3 said:
I know how to use the chain rule when I have variables. But it's when I'm dealing purely with operators I struggle. I.e

\frac{ \partial}{\partial u}

How can you apply a chain rule on this guy?
The definitions u=x-ct, v=x+ct imply x=(u+v)/2, t=(v-u)/2c, so

\frac{\partial}{\partial u}g(x(u,v),t(u,v))=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial g}{\partial t}\frac{\partial t}{\partial u}=\frac{\partial g}{\partial x}\frac{1}{2}+\frac{\partial g}{\partial t}\left(-\frac{1}{2c}\right)

and therefore

\frac{\partial}{\partial u}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{c}\frac{\partial}{\partial t}\right)
 
Back
Top