Show that the gradient operator is a vector operator.

[andresordonez]

Homework Statement

Hi, this is problem 2.5 from "Atomic Many Body Theory" - Lindgren, Morrison

An operator which transforms under a rotation in the same way as the vector $$\vec{r}$$ (or any other vector) is called a vector operator. Show that the gradient operator $$\vec{\nabla}$$ satisfies this condition for a rotation of 90 degrees about the z axis, or, in other words that

$$\nabla_x^{'} = P_z(90) \nabla_x P_z(-90) = \nabla_y$$

$$\nabla_y^{'} = -\nabla_x$$

$$\nabla_z^{'} = \nabla_z$$

Homework Equations

The operator which corresponds to a rotation by an angle $$\theta$$ about the z axis is
$$P_z(\theta) = \exp (-i \theta j_z)$$ where j is a generalized momentum

$$\nabla_x = i p_x$$ and the same is true for the y and z components

$$\l_z = x p_y - y p_x$$ where l is the orbital angular momentum operator

The Attempt at a Solution

I'm having problems with the 1st and 2nd equalities. This is how I did the third
$$\nabla_{z}^{'}=\exp\left\{ -i\alpha\left(xp_{y}-yp_{x}\right)\right\} ip_{z}\exp\left\{ i\alpha\left(xp_{y}-yp_{x}\right)\right\}$$

$$[p_{z},xp_{y}-yp_{x}]=[p_{z},xp_{y}]-[p_{z},yp_{x}]=x[p_{z},p_{y}]+[p_{z},y]p_{x}-y[p_{z},p_{x}]-[p_{z},y]p_{x}=0$$

$$\Rightarrow\nabla_{z}^{'}=i\exp\left\{ -i\alpha\left(xp_{y}-yp_{y}\right)\right\} \exp\left\{ i\alpha\left(xp_{y}-yp_{y}\right)\right\} p_{z}=ip_{z}=\nabla_{z}$$

I think this one is right, but if you see any error please let me know


This is one of my attempts at solving the third inequality (which is very similar to the first)

$$\nabla_{y}^{'}=-\nabla_{x}$$

$$\nabla_{y}^{'}=P_{z}(90\textdegree)\nabla_{y}P_{z}(-90\textdegree)$$

$$=\exp\left\{ -i\alpha\left(xp_{y}-yp_{x}\right)\right\} ip_{y}\exp\left\{ i\alpha\left(xp_{y}-yp_{x}\right)\right\}$$

$$=i\left\{ \sum_{n=0}^{\infty}\frac{(-i\alpha)^{n}}{n!}\left(xp_{y}-yp_{x}\right)^{n}\right\} p_{y}\left\{ \sum_{m=0}^{\infty}\frac{(i\alpha)^{m}}{m!}\left(xp_{y}-yp_{x}\right)^{m}\right\}$$

$$=i\left\{ \sum_{n=0}^{\infty}\frac{(-i\alpha)^{n}}{n!}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(xp_{y})^{n-k}(yp_{x})^{k}\right\} p_{y}\left\{ \sum_{m=0}^{\infty}\frac{(i\alpha)^{m}}{m!}\sum_{s=0}^{m}(-1)^{s}\binom{m}{s}(xp_{y})^{m-s}(yp_{x})^{s}\right\}$$

$$=i\sum_{n=0}^{\infty}\sum_{k=0}^{n}\sum_{m=0}^{\infty}\sum_{s=0}^{m}\frac{(-1)^{n}(i\alpha)^{n}}{n!}\frac{(i\alpha)^{m}}{m!}(-1)^{k}(-1)^{s}\binom{n}{k}\binom{m}{s}(xp_{y})^{n-k}(yp_{x})^{k}p_{y}(xp_{y})^{m-s}(yp_{x})^{s}$$

$$=i\sum_{n=0}^{\infty}\sum_{k=0}^{n}\sum_{m=0}^{\infty}\sum_{s=0}^{m}\frac{(i\alpha)^{n+m}}{n!m!}(-1)^{n+k+s}\binom{n}{k}\binom{m}{s}(xp_{y})^{n-k}(yp_{x})^{k}p_{y}(xp_{y})^{m-s}(yp_{x})^{s}$$

$$=i\sum_{n=0}^{\infty}\sum_{k=0}^{n}\sum_{m=0}^{\infty}\sum_{s=0}^{m}\frac{(i\alpha)^{n+m}}{n!m!}(-1)^{n+k+s}\binom{n}{k}\binom{m}{s}x^{n-k}p_{y}{}^{n-k}y^{k}p_{x}{}^{k}p_{y}x^{m-s}p_{y}{}^{m-s}y^{s}p_{x}{}^{s}$$

$$=i\sum_{n=0}^{\infty}\sum_{k=0}^{n}\sum_{m=0}^{\infty}\sum_{s=0}^{m}\frac{(i\alpha)^{n+m}}{n!m!}(-1)^{n+k+s}\binom{n}{k}\binom{m}{s}x^{n-k}p_{x}{}^{k}x^{m-s}p_{x}{}^{s}p_{y}{}^{n-k}y^{k}p_{y}p_{y}{}^{m-s}y^{s}$$

I've tried some other things but I don't think I'm getting any closer to the answer. Any help is much appreciated.

 

[mark726]

Thank you for your post. Let's take a look at your attempt at solving the third equality. It seems that you are on the right track, but there are a few mistakes in your calculations. I will point them out and provide some guidance on how to correct them.

First, let's look at the expression you have for \nabla_{y}^{'}. You correctly used the definition of the rotation operator P_{z}(\theta), but then you wrote it as \exp\left\{ -i\alpha\left(xp_{y}-yp_{x}\right)\right\}. This is not quite correct. Remember that j_{z} = xp_{y}-yp_{x}, so the correct expression should be \exp\left\{ -i\alpha j_{z}\right\}.

Next, you wrote p_{y} as the generalized momentum operator, which is correct. However, when you apply the commutation relation [p_{z},xp_{y}-yp_{x}] = 0, you wrote p_{y} as the physical momentum operator. This is incorrect. Remember that p_{y} is a part of the generalized momentum operator j_{z}, so you should use [p_{z},j_{z}] = 0 instead.

Finally, when you expand the exponential terms, you should use the binomial theorem to factor out the powers of x and y. This will simplify your expression significantly.

I hope this helps you to correct your mistakes and continue with your solution. If you need further assistance, please do not hesitate to ask. Good luck!