# Showing Complex Vectors are Orthonormal

1. Apr 15, 2017

### Crush1986

1. The problem statement, all variables and given/known data
let $$\epsilon_1$$ and $$\epsilon_2$$ be unit vectors in R3. Define two complex unit vectors as follows:
$$\epsilon_{\pm} = \frac{1}{\sqrt{2}}(\epsilon_1 \pm i \epsilon_2)$$

verify that epsilon plus minus constitutes a set of complex orthonormal unit vectors. That is, show that $$(\epsilon_\pm)^* \cdotp \epsilon_\mp = 0$$

2. Relevant equations
Dot Product.

3. The attempt at a solution
So... I don't know what I can possibly be missing. I do the dot product say of $$(\epsilon_+)^* \cdotp i \epsilon_-$$
and I'm ending up with, $$\frac{1}{2} [1-i \epsilon_2 \cdotp \epsilon_1-i \epsilon_2 \cdotp \epsilon_1-1]$$
So the complex parts don't go away? I'd appreciate any help... Thanks.

2. Apr 15, 2017

### PeroK

The complex inner product is not linear in both arguments.

3. Apr 15, 2017

### Crush1986

I'm sorry, I don't follow exactly.

4. Apr 15, 2017

### PeroK

5. Apr 15, 2017

### Crush1986

Hrm, ok I see. When you take the dot product one of the vectors must be changed to it's complex conjugate? I still must be doing something wrong as I'm not getting the advertised result.

6. Apr 15, 2017

### PeroK

In a complex vector space, you have:

$\langle \alpha u, v \rangle = \alpha^* \langle u, v \rangle$

$\langle u, \alpha v \rangle = \alpha \langle u, v \rangle$

This is the physicist's version. In pure mathematics, it's the other way round (like in the link I gave you). You need to know which approach is being used in your book or course.

7. Apr 15, 2017

### Crush1986

Oh no... I'm sorry I have a typo. I think that is what could be some of my confusion.

I'm trying $$\epsilon_+^* \cdotp \epsilon_-$$ not $$\epsilon_+^* \cdotp i \epsilon_-$$

Sorry if that has wasted your time. I've definitely learned something about complex dot products but I still haven't been able to resolve this unfortunately.

That first dot product is supposed to be zero...

8. Apr 15, 2017

### PeroK

You may be trying to take too many shortcuts. Write it out in full, starting with:

$\langle \epsilon_+, \epsilon_- \rangle = \langle \frac{1}{\sqrt{2}}(\epsilon_1 + i \epsilon_2), \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2) \rangle$

9. Apr 15, 2017

### Crush1986

yes I have that, except that the epsilon plus is complex conjugate of itself. so $$\langle \epsilon_+^*, \epsilon_- \rangle = \langle \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2), \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2) \rangle$$

Is that not correct? Then I get the cross terms don't cancel? and if the i for the epsilon - is positive I get 1? I may just not be getting how to do a complex dot product. Funny we didn't go over this, perhaps we did it in our math methods last year... but I don't remember doing them at all if we did.

10. Apr 15, 2017

### PeroK

You don't take the complex conjugate in that way. The inner product of two vectors is:

$\langle u, v \rangle$

What you are doing is:

$\langle u^*, v \rangle$

Which is the inner product of $u^*$ and $v$.

You may be thinking of the inner product expressed as a product of coefficients:

$\langle u, v \rangle = u^*_1v_1 + u^*_2v_2 + u^*_3 v_3$

11. Apr 15, 2017

### Crush1986

The problem says to show that the complex conjugate of epsilon plus dotted with epsilon minus is 0 though. I'm just not seeing any possible way the real and complex can disappear... I've tried this so many different times trying different things...

12. Apr 15, 2017

### PeroK

This is wrong. You need to show that $\epsilon_+ \cdotp \epsilon_- = 0$ and $\epsilon_\pm \cdotp \epsilon_\pm = 1$

Perhaps whoever set the question got confused.

Sorry, I didn't notice that this was part of the question, not your approach.

13. Apr 15, 2017

### Crush1986

Ok, I'll try that after some sleep. I was beginning to think there was an issue or typo perhaps! Thanks!