0
votes
4answers
58 views

Is raising a value to the second power the same as multiplying it it's complex conjugate?

I was watching a video on YouTube on Quantum Mechanics Concepts and saw that if you wanted to convert a probability amplitude to a probability, you square it. In the video he said that this was ...
1
vote
1answer
41 views

Quantum Fourier Transform and roots of unity.

I need to find $QFT_{6}$ for the state quantum state $\frac{1}{\sqrt2}(|0\rangle + |3\rangle)$. I received a very sufficient answer recently on simplifying nth roots of unity, but I am having a lot of ...
0
votes
2answers
88 views

How does one simplify exponents for complex primitive nth roots of unity?

Let us define a complex primitive N-th root of unity, omega: $$ \omega = \cos(\theta) + i\sin(\theta) \\ = e^{\frac{2\pi}{N}} $$ By the definition of an nth root of unity, ω is the second solution to ...
2
votes
1answer
196 views

Question on complex number calculation for transmission coefficient of finite potential well

This is actually in my quantum mechanics textbook (pure math question though), and I just cannot see why this equality is true. Any help would be greatly appreciated! Let $F$ and $A$ be nonzero ...
1
vote
1answer
42 views

Unwanted $i$ floating around when trying to calculate $\langle p\rangle$

$\def\sp#1{\left\langle#1\right\rangle}$I am given $$ \Psi(x,0)=A_0 \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \cdot \exp\left(\frac{i}{\hbar}p_0x\right)\tag1$$ where $A_0=(\pi ...
1
vote
0answers
195 views

The general recipe for finding the conjugate of a complex function

I have the general recipe for finding the complex conjugate of a function down as follows: Suppose I have $f(z)$: Separate $f(z)$ into a sum of real and imaginary functions such that ...
1
vote
1answer
211 views

Why are the coefficients of the base states of a qubit complex numbers?

Why are qubits represented as $$\left|{q}\right\rangle = \alpha\left|{0}\right\rangle+\beta\left|{1}\right\rangle\equiv\alpha\left[{1 \ 0}\right]^T+\beta\left[{0 \ 1}\right]^T; ...