Why psi squared instead of psi?

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Discussion Overview

The discussion centers around the rationale for using the absolute square of the wave function, ##|\psi|^2##, to derive probabilities in quantum mechanics, as opposed to simply squaring the real part of the wave function. It explores theoretical implications and interpretations related to the Schrödinger equation.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions why the absolute square of the wave function is used for probabilities instead of just squaring the real part to avoid imaginary values.
  • Another participant argues that the conservation of the absolute square of the wave function can be derived from the Schrödinger equation, while the real or imaginary parts do not have this property when altered by a global phase.
  • A further contribution highlights that the interpretation of ##|\psi(t,\vec{x})|^2## as a probability distribution is contingent on the constancy of the integral of the absolute square over space, as dictated by the Schrödinger equation, and that ##|\psi|^2 \geq 0## is essential for a valid probability distribution.
  • A link to additional resources is provided for further reading on the topic.

Areas of Agreement / Disagreement

Participants present differing views on the appropriateness of using the absolute square versus the real part of the wave function, indicating that multiple competing perspectives remain without a consensus.

Contextual Notes

The discussion does not resolve the underlying assumptions regarding the nature of the wave function or the implications of different interpretations of quantum mechanics.

kent davidge
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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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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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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.
 
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