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If i have equation:

\begin{align} P = \left|\psi\right|^2 \end{align}

where $P$ is a probability and we know there is no negative probability. This means $P$ must belong to $\mathbb{R}$. If i want to calculate $|\psi|$ i can do it simply by sq. rt. the equation:

\begin{align} \left|\psi\right| = \sqrt{P} \end{align}

Is $\left|\psi\right|$ a real number? What about $\psi$ by itself? Please explain.

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You cannot get $\psi$ from $P$, only its magnitude can be found. Even with real numbers you cannot find a unique $x$ from $x^2=9$, you only get the absolute value $|x|=3$. The phase angle of $\psi$ cannot be recovered from the given equation. – Maesumi May 11 '13 at 11:41
The modulus $|z|$ is always a positive real number, and its square will also be nonnegative. – rschwieb May 11 '13 at 12:22
But there is a possibility that $\psi$ is complex? – 71GA May 11 '13 at 12:49
Yes, that is a possibility. – Harald Hanche-Olsen May 11 '13 at 12:51
But it could allso be real right? – 71GA May 11 '13 at 13:01
up vote 2 down vote accepted

The absolute value of a complex number (sometimes called its modulus) is real by construction. If $\psi=\alpha+i\beta$ with $\alpha$ and $\beta$ real, then $|\psi|=\sqrt{\alpha^2+\beta^2}$.

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This means $\left|\psi\right|$ is always real and $\psi$ complex? Can you please edit your anwser and show me how you got equation $\left|\psi\right|=\sqrt{\alpha^2+\beta^2}$. – 71GA May 11 '13 at 11:32
That equation is the definition of the absolute value. So I can't show you how I “got” it, except perhaps by pointing to a thousand textbooks that all define it that way. So yes, that means $|\psi|$ is real. The square root is the usual square root as applied to a positive real numbers. – Harald Hanche-Olsen May 11 '13 at 11:36
I'm nitpicking, I know, but it's not only real, it's even guaranteed to be $\geq 0$. – fgp May 11 '13 at 11:53
@fgp: The modulus is a wonderful beast indeed! – Harald Hanche-Olsen May 11 '13 at 11:55
@HaraldHanche-Olsen It's nothing compared to the phase! Oh, the pure joy of recklessly adding $2\pi$ and watching nothing change... always shuddering at bit at the dreaded possibility that I, one day, just might happen to try this on a riemannian manifold.... – fgp May 11 '13 at 12:01

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