A more simple question, so I am watching a quantum mechanics lecture on potentials of free particles and am doing the general solution of schrodinger's stationary equation for a free particle when I was told to normalize the solution (which I can do all well and good) but I have no idea what it actually means to "normalize" My question being what is normalization ? What does its product describe ?
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4You can multiply a wave-function $\psi$ by any constant and it will still describe the same physical state. However, the Born rule tells us that the square $|\psi|^2$ corresponds to a probability. And so it follows that the integral $\int |\psi|^2$ over all possible states must be 1. Normalisation is just scaling $\psi$ by a constant to make sure this integral is indeed unity. – lemon Mar 06 '16 at 17:02
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@lemon, shouldn't that be an answer? – Alfred Centauri Mar 06 '16 at 17:06
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what is meant by an integral being unity ? – darren Mar 06 '16 at 17:06
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@darren "unity" is a (uselessly complicated) way that a lot of physicists say "one". When they say something is unity, they mean that thing is equal to one. – DanielSank Mar 06 '16 at 17:11
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Related: Who is doing the normalization of wave function in the time evolution of wave function? – ACuriousMind Mar 06 '16 at 17:15
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so what does this show if the integral is unity ? – darren Mar 06 '16 at 17:26
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@darren That the square $|\psi|^2$ can be interpreted as a probability. – lemon Mar 06 '16 at 18:19
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Born's rule: the probability density of finding a particle in a certain place is proportional to its square absolute value.
To change the "is proportional to" to "is", you multiply the wave function by a constant so that the absolute value squared integrates to 1, and so acts as a probability density function.
That's called normalisation, or normalising the wave function.
Andrea
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One peculiar fact about a real life wave function $\psi$ is that it can be normalized. In order to analyze and compare the various outcomes of the solution of a Schrodinger equation, one need to assign a quality that is unique to all the wave functions, that which is to transform them such that their area is always 1.
Sathyam
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1Wrong, if you always divide by the norms in the probability amplitude, you don't need to normalize at all. – ACuriousMind Mar 06 '16 at 17:15
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@ACuriousMind True, still the statement is valid, whether you know the norm or not. – Sathyam Mar 06 '16 at 17:17
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@AccidentalFourierTransform Solution to potentials that don't have an infinity in a local space. – Sathyam Mar 06 '16 at 17:18