I Why psi squared instead of psi?

kent davidge

Why should we square the abs value of the wave function to get probabilities? Why don't just forget the imaginary part, and (in order to get positive values) square the real part of the function?

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Truecrimson

If you take the Schrödinger equation as given, you can derive that the absolute square of the wave function is conserved but not the square of the real or imaginary part. (You can move around the real and imaginary parts by multiplying the wave function by a global phase $e^{i \phi}$ without changing anything physical.)

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vanhees71

Gold Member
The possibility to interpret $|\psi(t,\vec{x})|^2$ as a probability distribution hinges on the fact that
$$\|\psi \|^2=\int_{\mathrm{R}^3} |\psi(t,\vec{x})|^2=\text{const}$$
due to the Schrödinger equation,
$$\mathrm{i} \partial_t \psi(t,\vec{x})=-\frac{\Delta}{2m} \psi(t,\vec{x}) + V(\vec{x}) \psi(t,\vec{x}).$$
Of course, also $|\psi|^2 \geq 0$ is also important for having a probability distribution.

kent davidge

"Why psi squared instead of psi?"

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