Normalisable wavefunction

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In quantum mechanics, wave functions which describe real particles must be normalisable¹: the probability of the particle to occupy any place must equal 1. Mathematically, in one dimension this is expressed as

$\int_{A}^{B} \psi^*(x)\psi(x)\ dx=1$

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

All wavefunctions which represent real particles must be normalizable, 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 Schrödinger equation for certain boundary conditions to discard solutions which do not have a finite integral at a given interval. For example, this disqualifies periodic functions as wave function solutions for infinite intervals, while those functions can be solutions for finite intervals.

1 Derivation of normalisation
2 Example of normalisation
3 Proof that wavefunction normalisation doesn't change associated properties
4 Note
5 See also

[edit] Derivation of normalisation

In general, $ψ$ is a complex function. However,

$\psi^* \psi = \mid \psi \mid ^2$

is real, greater than zero, and is known as a probability density function.

This means that

$p(-\infty \le x \le \infty) = \int_{-\infty}^{\infty} \mid \psi \mid ^2 dx. \quad (1)$

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

$p(-\infty \le x \le \infty) = \int_{-\infty}^{\infty} \mid \psi \mid ^2 dx = 1. \quad (2)$

If the integral is finite, we can multiply the wavefunction, $ψ$ , by a constant such that the integral is equal to 1. Alternatively, if the wavefunction already contains an appropriate arbitrary constant, we can solve equation (2) to find the value of this constant which normalises the wavefunction.

[edit] Example of normalisation

A particle is restricted to a 1D region between x=0 and x=l; its wavefunction is:

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

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

$\int_{-\infty}^{\infty} \mid \psi \mid ^2 dx = 1$

to find A.

Substituting $ψ$ into $\mid \psi \mid ^2$ we get

$\mid \psi \mid ^2 = A^2 e^{i(kx - \omega t)} e^{-i(kx - \omega t)} =A^2$

$\int_{-\infty}^{0} 0 dx + \int_{0}^{l} A^2 dx + \int_{l}^{\infty} 0 dx = 1$

therefore

$A^2 l = 1 \Rightarrow A = \left ( \frac{1}{\sqrt{l}} \right ).$

The normalised wavefunction is:

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

[edit] Proof that wavefunction normalisation doesn't change associated properties

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

$\frac{-\hbar}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) = E \psi (x).$

If $ψ$ is normalised and replaced with $A ψ$ , then

$\frac{d(A \psi)}{d x} = A \frac{d \psi}{d x}$ and $\frac{d^2 (A \psi)}{d x^2} = \frac{d}{d x} \left ( \frac{d (A \psi)}{d x} \right ) = A \frac{d^2 \psi}{d x^2}.$

The Schrödinger wave equation therefore becomes:

$\frac{-\hbar}{2m} A\frac{d^2 \psi}{d x^2} + V(x) A \psi (x) = E A \psi(x)$

$\Rightarrow A \left ( \frac{-\hbar}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) \right ) = A \left ( E \psi (x) \right )$

$\Rightarrow \frac{-\hbar}{2m} \frac{d^2 \psi}{d x^2} + V(x) \psi (x) = E \psi (x)$

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

[edit] Note

Note 1: The spelling normalisable is a British variant spelling of normalizable.

Category: Quantum mechanics

Normalisable wavefunction

From Wikipedia, the free encyclopedia

Contents

[edit] Derivation of normalisation

[edit] Example of normalisation

[edit] Proof that wavefunction normalisation doesn't change associated properties

[edit] Note

[edit] See also

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