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

Question

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 equivalent to multiply by it's complex conjugate. So is this correct? Is squaring the same as multiplying by the complex conjugate, or is this just a thing you do in Quantum Mechanics? Also, I'm not sure if this should be asked on the physics.se instead of math.se, if it should then I'm sorry.

Answer 1

A probability is a real number (between $0$ and $1$). The magnitude of a complex number $z$ is $|z|$, which is real. Now $z \bar z = |z|^2$, so multiplying by the complex conjugate gives the square of its magnitude, not the square of the complex number itself.

For example, $i^2 = -1$ but $|i|^2 = 1$.

Answer 2

No, it is not: Given $i$, we have

$$i^2 = -1 \ne 1 = i \overline{i}$$

What we can say, however, is that the absolute value of any complex $z$ is defined to be

$$|z| = \sqrt{z \overline{z}}$$

Answer 3

No, they're different, in both math and quantum mechanics. What is true, however, is that $z\bar z=|z|^2$. In QM, the probability is the square of the absolute value of the amplitude.

Answer 4

Not exactly. You don't square the complex number itself, you square its magnitude. You can see this is true by actually doing the computation for each value: Let $z=a+bi$. First the modulus: $$\left|z\right|^2=\sqrt{a^2+b^2}^2=a^2+b^2$$ Then the conjugate: $$z\bar{z}=(a+bi)(a-bi)=a^2-abi+abi-b^2i^2=a^2-(-1)b^2=a^2+b^2$$

So, yes, squaring the magnitude is the same thing as multiplying by the conjugate.