Operator and Commutor Question

[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 $$x$$.

 

[Cthugha]

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 $$A_{new} = e^{i \gamma} A_{old} e^{i \gamma}$$ ?

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.

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 $$x$$.

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.

 

[Domnu]

Thanks a bunch =]