Showing Complex Vectors are Orthonormal

[Crush1986]

Homework Statement

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

Homework Equations

Dot Product.

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.

 

[PeroK]

The complex inner product is not linear in both arguments.

 

[Crush1986]

The complex inner product is not linear in both arguments.

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

 

[PeroK]

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

http://mathworld.wolfram.com/HermitianInnerProduct.html

 

[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.

 

[PeroK]

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.

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.

 

[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...

 

[PeroK]

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...

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

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

 

[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.

 

[PeroK]

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.

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:

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

 

[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...

 

[PeroK]

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

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.

 

[Crush1986]

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.

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

 

[mdpugh]

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.

If $\varepsilon_1$ and $\varepsilon_2$ are orthogonal, the cross terms vanish. Although it isn't mentioned, I believe they need to be orthonormal vectors.