# Operator and Commutor Question

1. Jun 19, 2008

### Domnu

In this site: http://farside.ph.utexas.edu/teaching/qm/lectures/node23.html , how does one go from steps 141, 142 to 143? Shouldn't the result in 143 be $$A_{new} = e^{i \gamma} A_{old} e^{i \gamma}$$ ? Also, how does step 144 work (the entire thing from left to right?)

For step 144, here's how I see it (assuming step 143 is correct):

$$e^{-i \gamma} \frac{d}{dx} e^{i \gamma} = i e^{-i \gamma} \frac{d \gamma}{dx} e^{i \gamma} = i e^{-i \gamma} e^{i \gamma} \frac{d \gamma}{dx} = i \frac{d \gamma}{dx}$$

in other words, I don't see how the extra $$d/dx$$ term popped up. Also, is my above work correct? I effectively stated that $$d\gamma/dx$$ and $$e^{i \gamma}$$ commuted as operators since they're both functions of $$x$$.

2. Jun 19, 2008

### Cthugha

You want $$\langle_{new}A_{new}\rangle_{new}$$ to be the same as $$\langle_{old}A_{old}\rangle_{old}$$.
By putting in the old bras and kets for the new ones, you get:
$$\langle_{old} e^{i \gamma}A_{new} e^{- i \gamma}\rangle_{old}$$.
In order to compensate these additional terms, you have to replace $$A_{new}$$ by $$e^{-i \gamma}A_{old} e^{i \gamma}$$ as stated in your reference.

Remember, that you are transforming an operator. Operators usually work on some kind of function. Just imagine, that it works on some function f(x) and it will get much clearer. Then:

$$e^{-i \gamma} \frac{d}{dx} e^{i \gamma} f(x) = i e^{-i \gamma} \frac{d \gamma}{dx} e^{i \gamma} f(x) + \frac{d f(x)}{dx}=i \frac{d \gamma}{dx} f(x)+\frac{d f(x)}{dx}$$

So it is more or less the derivative of a product, which you have to keep in mind.

3. Jun 19, 2008

### Domnu

Thanks a bunch =]