Cross product - non commutative algebra

In quantum mechanics I am told that the cross product of angular momentum with itself is not zero and this is because r and momentum (p) do not commute.

I have $\displaystyle \underline l = \underline r \times \underline p$ and $\displaystyle \underline p = (h/i) \nabla $and I want to calculate l cross l.

I cannot write $\displaystyle \underline l \times \underline l = \hat x_i \epsilon_{ijk}r_jp_k$ because this definition of the cross product requires the algebra to be commutative

Similarly I cannot use the determinant mnemonic for the cross product because it makes the same assumption?

So how do I show that:

$\displaystyle \underline l \times \underline l = ih \underline l $?

Re: Cross product - non commutative algebra

Hey Kiwi_Dave.

Can you expand the definition by substituting in p = (h/i)del so you get l x l = (r x (h/i)del) x (r x (h/i)del) = r x -l x (h/i)del = -p x (h/i)del. -(h/i)*i/i = ih so you combine that to get ih(p x del). But what is the definition of angular momentum?

(My physics is knowledge is very limited, but hopefully this can help you solve the problem).

Re: Cross product - non commutative algebra

Solved: Cross product - non commutative algebra

$\displaystyle \underline l \times \underline l = (\underline r \times \underline p) \times (\underline r \times \underline p) = \underline a \times \underline b$ where:

$\displaystyle \underline b = (yP_z-zP_y)\hat x + (zP_x-xP_z)\hat y +(xP_y-yP_x)\hat z$ and

$\displaystyle \underline a = \frac hi \left[(y\frac \partial {\partial z}-z\frac \partial {\partial y})\hat x + (z\frac \partial {\partial x}-x\frac \partial {\partial z})\hat y +(x\frac \partial {\partial y}-y\frac \partial {\partial x})\hat z \right]$

So

$\displaystyle \underline l \times \underline l = \frac hi \left| \begin{array}{ccc} \hat x & \hat y & \hat z \\ (y\frac \partial {\partial z}-z\frac \partial {\partial y}) & (z\frac \partial {\partial x}-x\frac \partial {\partial z}) & (x\frac \partial {\partial y}-y\frac \partial {\partial x}) \\ (yP_z-zP_y) & (zP_x-xP_z) & (xP_y-yP_x)\hat z \end{array} \right|$ So:

$\displaystyle \underline l \times \underline l = \frac hi [(zP_y-yP_z) \hat x - (zP_x-xP_z) \hat y + (yP_x-xP_y) \hat z]$

$\displaystyle \underline l \times \underline l = \frac hi \left| \begin{array}{ccc} \hat x & \hat y & \hat z \\ P_x & P_y & P_z \\ x & y & z \end{array} \right| =ih \underline l$

Re: Cross product - non commutative algebra

Quote:

Originally Posted by

**mathlover10** is this from Griffiths?

No, I am studying Noether's theorem (self study) and it delves into many areas of Physics.

Re: Cross product - non commutative algebra

After a large amount of tedious work I have to say that I disagree with your final solution that $\displaystyle \overline{L} \times \overline{L} = ih \overline{L}$.

I'll give you the relevant steps and leave it to you to work through the details to see if I have made a mistake.

First:

$\displaystyle \overline{L} \times \overline{L} \propto \left ( \overline{r} \times \overline{ \nabla } \right ) \times \left ( \overline{r} \times \overline{ \nabla } \right )$

Since the cross product is associative we may regroup this as:

$\displaystyle = \left \[ \overline{r} \times \left ( \overline{ \nabla } \times \overline{r} \right ) \right \] \times \overline{ \nabla } $

The term in the brackets is known so I'll just quote it:

$\displaystyle = \left \[ \frac{1}{2} \overline{ \nabla } \left ( \overline{r} \cdot \overline{r} \right ) - \left ( \overline{r} \cdot \overline{ \nabla } \right ) \overline{r} \right \] \times \overline{ \nabla }$

Let's look at each term separately:

$\displaystyle \frac{1}{2} \overline{ \nabla } \left ( \overline{r} \cdot \overline{r} \right ) = \left ( x \hat{i} + y \hat{j} + z \hat{k} \right )$

So

$\displaystyle \left \[ \frac{1}{2} \overline{ \nabla } \left ( \overline{r} \cdot \overline{r} \right ) \right \] \times \overline{ \nabla } = \left ( x \hat{i} + y \hat{j} + z \hat{k} \right ) \times \overline{ \nabla }$

$\displaystyle = L_x \hat{i} - L_y \hat{j} + L_z \hat{k}$

Now for the second term:

$\displaystyle \left \[ \left ( \overline{r} \cdot \overline{ \nabla } \right ) \overline{r} \right \] \times \overline{ \nabla }$

$\displaystyle = \left \[ \left ( x \frac{ \partial }{ \partial x } + y \frac{ \partial }{ \partial y } + z \frac{ \partial }{ \partial z} \right ) \overline{r} \right \] \times \overline{ \nabla }$

$\displaystyle = L_x \hat{i} - L_y \hat{j} + L_z \hat{k}$

Finishing up we get:

$\displaystyle = \left \[ \overline{r} \times \left ( \overline{ \nabla } \times \overline{r} \right ) \right \] \times \overline{ \nabla } = \left ( L_x \hat{i} - L_y \hat{j} + L_z \hat{k} \right ) - \left ( L_x \hat{i} - L_y \hat{j} + L_z \hat{k} \right )$

Thus

$\displaystyle \overline{L} \times \overline{L} = 0$

There is a more physical basis of this result. $\displaystyle \overline{L} = m \overline{r} \times \overline{p}$ is a pseudovector. That means that $\displaystyle \overline{L} \times \overline{L}$ is a vector. But if we have $\displaystyle \overline{L} \times \overline{L} = ih \overline{L}$ then we are equating a vector with a pseudovector, a contradiction. Thus $\displaystyle \overline{L} \times \overline{L} = 0$.

-Dan

Re: Cross product - non commutative algebra

Quote:

Originally Posted by

**topsquark** There is a more physical basis of this result. $\displaystyle \overline{L} = m \overline{r} \times \overline{p}$ is a pseudovector. That means that $\displaystyle \overline{L} \times \overline{L}$ is a vector. But if we have $\displaystyle \overline{L} \times \overline{L} = ih \overline{L}$ then we are equating a vector with a pseudovector, a contradiction. Thus $\displaystyle \overline{L} \times \overline{L} = 0$.

Please disregard this statement. It is incorrect. (Doh) That's what I get for staying up and doing Physics at 3 in the morning!

-Dan

Re: Cross product - non commutative algebra

Thanks Dan

First up I am getting malware warnings when visiting MHF. Google Crome suddenly refuses to come here.

I have two "concerns" with your working. Probably just my understanding:

1. You say that the cross product is associative. I can't find that written anywhere but find sites (like Cross Product Not Associative - ProofWiki) that say it is not associative.

2. You have used an identity that I worry may only be valid when the multiplication commutes?

The text I am working from says l cross l = ihl without any working, he just stated it as fact. That said I have found so many errors in this particular text that I don't trust anything that I cannot prove for myself (otherwise it is a great text).

Re: Cross product - non commutative algebra

Quote:

Originally Posted by

**Kiwi_Dave** I have two "concerns" with your working. Probably just my understanding:

1. You say that the cross product is associative. I can't find that written anywhere but find sites (like

Cross Product Not Associative - ProofWiki) that say it is not associative.

2. You have used an identity that I worry may only be valid when the multiplication commutes?

The text I am working from says l cross l = ihl without any working, he just stated it as fact. That said I have found so many errors in this particular text that I don't trust anything that I cannot prove for myself (otherwise it is a great text).

Oh dear. I never knew cross products weren't associative. (Crying) You learn something new every day, I guess.

Well, I guess that takes care of my derivation! Hmmm...I wonder if the assumption of associativity would imply commutivity? I'll have to look that one up. Thanks for the catch!

-Dan

Re: Cross product - non commutative algebra

Quote:

Originally Posted by

**Kiwi_Dave** First up I am getting malware warnings when visiting MHF. Google Crome suddenly refuses to come here.

I know, I've had to turn off the malware warning in Chrome and trust Norton 360 to safeguard my computer. I don't know what's going on yet, nor how to fix it. Thanks for letting me know.

-Dan