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#REDIRECT ] {{R from merge}}
In ], ]s which describe real ]s must be '''normalisable'''{{fn|1}}: the ] of the particle to occupy any place must equal 1. ], in one ] this is expressed as

:<math>\int_{A}^{B} \psi^*(x)\psi(x)\ dx=1</math>

in which the integration parameters ''A'' and ''B'' indicate the interval in which the particle must exist.

All wave functions which represent real particles must be normalisable, that is, they must have a total probability of one - they must describe the probability of the particle existing as 100%. This trait enables anyone who solves the ] for certain boundary conditions to discard solutions which do not have a finite integral at a given interval. For example, this disqualifies ]s as wave function solutions for infinite intervals, while those functions can be solutions for finite intervals.

==Derivation of normalisation==
In general, <math>\psi</math> is a ] ]. However,

:<math>\psi^* \psi = \mid \psi \mid ^2</math>

is ], greater than or equal to zero, and is known as a ].

This means that

:<math>p(-\infty \le x \le \infty) = \int_{-\infty}^{\infty} \mid \psi \mid ^2 dx. \quad (1)</math>

where ''p''(''x'') is the probability of finding the particle at ''x''. Equation (1) is given by the definition of a ]. Since the particle exists, its probability of being anywhere in space must be equal to 1. Therefore we integrate over all space:

:<math>p(-\infty \le x \le \infty) = \int_{-\infty}^{\infty} \mid \psi \mid ^2 dx = 1. \quad (2)</math>

If the integral is finite, we can multiply the wave function, <math>\psi</math>, by a constant such that the integral is equal to 1. Alternatively, if the wave function already contains an appropriate arbitrary constant, we can solve equation (2) to find the value of this constant which normalises the wave function.

==Example of normalisation==
A particle is restricted to a 1D region between ''x''=0 and ''x''=l; its wave function is:

:<math>\psi (x,t) = \begin{cases} Ae^{i(kx-\omega t)}, \quad 0 \le x \le l \\ 0, \quad elsewhere. \end{cases}</math>

To normalise the wave function we need to find the value of the arbitrary constant ''A'', i.e., solve

:<math> \int_{-\infty}^{\infty} \mid \psi \mid ^2 dx = 1 </math>

to find ''A''.

Substituting <math>\psi</math> into <math> \mid \psi \mid ^2 </math> we get

:<math> \mid \psi \mid ^2 = A^2 e^{i(kx - \omega t)} e^{-i(kx - \omega t)} =A^2 </math>

so

:<math> \int_{-\infty}^{0} 0 dx + \int_{0}^{l} A^2 dx + \int_{l}^{\infty} 0 dx = 1 </math>

therefore

:<math>A^2 l = 1 \Rightarrow A = \left ( \frac{1}{\sqrt{l}} \right ).</math>

The normalised wave function is:

:<math> \psi (x,t) = \begin{cases} \left ( \frac{1}{\sqrt{l}} \right )e^{i(kx-\omega t)}, \quad 0 \le x \le l \\ 0, \quad \text{elsewhere.} \end{cases}</math>

==Proof that wave function normalisation doesn't change associated properties==

If normalisation of a wave function changed the properties associated with the wave function, the process becomes pointless as we still cannot yield any information about the properties of the particle associated with the un-normalised wave function. It is therefore important to establish that the properties associated with the wave function are not altered by normalisation.

All properties of the particle such as probability distribution, momentum, energy, expectation value of position etc. are derived from the Schrödinger wave equation. The properties are therefore unchanged if the Schrödinger wave equation is invariant under normalisation.

The Schrödinger wave equation is

:<math> \frac{-\hbar^2}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) = E \psi (x). </math>

If <math>\psi</math> is normalised and replaced with <math>A\psi</math>, then

:<math> \frac{d(A \psi)}{d x} = A \frac{d \psi}{d x}</math> and <math> \frac{d^2 (A \psi)}{d x^2} = \frac{d}{d x} \left ( \frac{d (A \psi)}{d x} \right ) = A \frac{d^2 \psi}{d x^2}.</math>

The Schrödinger wave equation therefore becomes:

:<math> \frac{-\hbar^2}{2m} A\frac{d^2 \psi}{d x^2} + V(x) A \psi (x) = E A \psi(x)</math>

:<math> \Rightarrow A \left ( \frac{-\hbar^2}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) \right ) = A \left ( E \psi (x) \right )</math>

:<math> \Rightarrow \frac{-\hbar^2}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) = E \psi (x) </math>

which is the original Schrödinger wave equation. That is to say, the Schrödinger wave equation is ] under normalisation, and consequently associated properties are unchanged.

==Note==
{{fnb|1}}The spelling ''normalizable'' is an ] of ''normalisable''.

== See also ==

* ]
* ]
* ]

==External links==



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