Problems with complex numbers and vectors

[Thadis]

Homework Statement

Prove the following statements about the inner product of two complex vectors with the same arbitrary numbers of components.

(a)<u|w>=<w|u>*
(b)|<u|w>|^2=|<w|u>|^2

Homework Equations

1. <u|w>=(u*)w
2. (c_1+c_2)*=c_1*+c_2*
3. c**=c
4. ((c_1)(c_2))*=(c_1*)c_2*

The Attempt at a Solution

I am having to do this for my Honors physics class in College and am in a Section of entry level Quantum Mechanics. Though I have yet been in a math class that has covered any sort of complex number math so I have been very lost when it comes to some of math that you do with complex numbers. I appologize if I should of posted this in a physics forum but I felt like since this is a math focused one it would be appropriate here.

For problems (a) I believe I have a solution done. What I basically did is use equation 1 to expand into a sum of n complex conjugate of u times n w's. After this then I used rule 2 and 3 to show that the two sides of the equation are equal.

For problem (b) though I have no idea even really how to do the squaring of the dot product. I assumed I would treat it like finding the inner product of the two vectors like in the previous problem and then squaring it but after that I have no idea what to do as the two sides are still not the same.

 

[Fredrik]

Let z be an arbitrary complex number. Denote its real and imaginary parts by x and y respectively. We have z=x+iy. The absolute value of z is the real number |z| defined by
$$|z|=\sqrt{x^2+y^2}.$$ The complex conjugate of z is the complex number z* defined by
$$z^*=x-iy.$$ You should use these definitions to prove the following theorem, and then use it to solve the problem.

For all complex numbers z, we have ##|z|^2=z^*z##.

 

[Thadis]

Would it be that you can set the individual components of the sum to a value of |z|^2 since the indivudual components would follow the form of (u_i*)w_i?

 

[Fredrik]

For problems (a) I believe I have a solution done. What I basically did is use equation 1 to expand into a sum of n complex conjugate of u times n w's. After this then I used rule 2 and 3 to show that the two sides of the equation are equal.

I prefer to prove once and for all that for all matrices A and B such that AB is defined, we have ##(AB)^*=B^*A^*##. (I'm not sure you even have to prove that here, since it's an elementary result of linear algebra). Then we can use that result to show (with a very short calculation) that ##\langle u|v\rangle=\langle v|u\rangle^*## for all ##u,v\in\mathbb C^n##. But your way is fine too. You are proving a special case of ##(AB)^*=B^*A^*## as a part of your calculation.

 

[Dick]

I prefer to prove once and for all that for all matrices A and B such that AB is defined, we have ##(AB)^*=B^*A^*##. (I'm not sure you even have to prove that here, since it's an elementary result of linear algebra). Then we can use that result to show (with a very short calculation) that ##\langle u|v\rangle=\langle v|u\rangle^*## for all ##u,v\in\mathbb C^n##. But your way is fine too. You are proving a special case of ##(AB)^*=B^*A^*## as a part of your calculation.

You should maybe say here you don't mean '*' to be just conjugation. You mean conjugation and transpose.

 

[Fredrik]

Would it be that you can set the individual components of the sum to a value of |z|^2 since the indivudual components would follow the form of (u_i*)w_i?

It's easier than that. You don't have to use the definition of matrix multiplication to evaluate u*v, not for part (b) anyway. (Not for part (a) either if you consider ##(AB)^*=B^*A^*## to be an elementary result from linear algebra that you're allowed to use).

Note that what I suggested that you should prove in post #2 doesn't have anything to do with the inner product on ##\mathbb C^n## that you're working with. It's just an elementary property of the absolute value function and the complex conjugate operation on ##\mathbb C##.

If you use the result I suggested that you prove, to evaluate the left-hand side of the equality in (b), what do you get? Do you get an expression with the property that if you were to swap u and w, the value of the expression would remain the same? If you do, then you're already done.

 

[Fredrik]

You should maybe say here you don't mean '*' to be just conjugation. You mean conjugation and transpose.

Yes, that's definitely worth mentioning. Physics texts often use the notation ##A^\dagger## instead of ##A^*## to force the readers to look at the definition instead of just assuming (incorrectly) that * denotes complex conjugation of the components.

 

[Thadis]

I think looking over a couple more questions I have I think mostly where I am getting confused is what to do with the |<u|w>|^2 operator even in the first place. I apogolize for not knowing more about this subject its just I have nowhere really to explain the details of a problem of this sort in more detail.

 

[Fredrik]

I'm not sure I understand what's confusing you. <u|w> is a complex number, so |<u,w>| is a real number (see the definition of the absolute value in post #2), and |<u,v>|² is the square of that real number. Have you tried proving the result I mentioned in post #2? Have you tried to use it the way I suggested in post #6?

 

[Thadis]

Oh wait I think I understand what now what it is. Basically for the second one what was my problem was I wasn't thinking about taking the conjugate of the two matrices multiplied together and multiplying it by the original matrix.

The reason why I am confused is just that we have just spent like 20 minutes in class going over this, which for the large majority of the class had no prior knowledge of either matrices or complex numbers. Reading yours and I believe someones elses explanations that talked about matrices are really what helped me just get the idea of the stuff better as I at least know the basics of Matrix Algebra.

 

[Fredrik]

If you haven't done the (easy) exercise I suggested in #2, you should, since it's so easy and one the most important results about complex numbers.

If you use the result of that exercise and the result of part (a), then part (b) is trivial. You will not even have to use the definition of this inner product (your eq. 1).

I'm going to bed, so no more posts tonight.

 

[Thadis]

Yeah I have done it and it also was explain in the physics textbook. After thinking about it and just hearing a couple of additional things from other places online I think I am making more sense of everything. Thanks again for all of the help!