Misplaced Pages

This is an old revision of this page, as edited by F=q(E+v^B) (talk | contribs) at 15:08, 21 January 2012 (extend scope). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

In quantum mechanics, wave functions are probability amplitudes containing information about the quantum system. The outcomes of measuring a quantum state, such as the location of a particle, are prescribed by the wavefunction, and since probabilities sum to 1 (one of the axioms), wavefunctions must be normalizable so the total probability of all outcomes is unity. The total is an integral over continuous variables like position, or sums over descrete variables like spin angular momentum.

For the case of the position of a particle, the total probability of finding the particle is 1 - if the particle is to exists somewhare. For given boundary conditions, this enables solutions to the Schrödinger equation to be discarded, if their integral diverges over the relavent interval. For example, this disqualifies periodic functions as wave function solutions for infinite intervals, while those functions can be solutions for finite intervals.

Probability and probability density

In general, Ψ is a complex function, it has no direct interpretation. However, the quantity

${\displaystyle \rho \psi ^{*}\psi =\mid \psi \mid ^{2}}$

is real, and positive definite (always greater than zero), and is the probability density function. This quantity has the interpretation of the probability the system is in a given state. Here, * (asterisk) indicates the complex conjugate.

Position normalization

For one particle in one dimension, the normalization condition is:

${\displaystyle \int _{-\infty }^{\infty }\left|\Psi (x,t)\right|^{2}dx=1}$

where the integration is in the interval (meaning "from ${\displaystyle \scriptstyle -\infty }$ to ${\displaystyle \scriptstyle \infty }$") indicates that the probability that the particle exists somewhere is unity.

For three dimensions, the integral is over all of space

${\displaystyle \iiint _{\rm {all\,space}}\left|\psi ({\mathbf {,}}t)\right|^{2}d^{3}{\mathbf {r}}=1}$

Derivation of normalization

This means that

${\displaystyle p(-\infty <x<\infty )=\int _{-\infty }^{\infty }\mid \psi \mid ^{2}dx.\quad (1)}$

where ${\displaystyle p(x)}$ is the probability of finding the particle at ${\displaystyle x}$. 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:

${\displaystyle p(-\infty <x<\infty )=\int _{-\infty }^{\infty }\mid \psi \mid ^{2}dx=1.\quad (2)}$

If the integral is finite, we can multiply the wave function, Ψ, 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 normalizes the wave function.

Plane-waves

Plane waves are normalized in a box or to a Dirac delta in the continuum approach. They are not normalizable over all space, since the integral doesn't converge.

Example of normalization

A particle is restricted to a 1D region between ${\displaystyle x=0}$ and ${\displaystyle x=l}$; its wave function is:

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

To normalize the wave function we need to find the value of the arbitrary constant ${\displaystyle A}$; i.e., solve

${\displaystyle \int _{-\infty }^{\infty }\mid \psi \mid ^{2}dx=1}$

to find ${\displaystyle A}$.

Substituting Ψ into ${\displaystyle \mid \psi \mid ^{2}}$ we get

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

so,

${\displaystyle \int _{-\infty }^{0}0dx+\int _{0}^{l}A^{2}dx+\int _{l}^{\infty }0dx=1}$

therefore;

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

Hence, the normalized wave function is:

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

Normalization invariance

It is important that the properties associated with the wave function are invariant under normalization. If normalization 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-normalized wave function.

All properties of the particle such as probability distribution, momentum, energy, expectation value of position etc.; are solved from the Schrödinger equation. Since the Schrödinger equation is linear, it is simple to see that the properties unchanged wave function is normalized.

The Schrödinger equation is:

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

If Ψ is normalized and replaced with ${\displaystyle A\psi }$, then the equation becomes:

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

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

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

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

See also

• Quantum mechanics
• Schrödinger equation
• Wave function

References

1. Griffiths, David J. (April 10, 2004). Introduction to Quantum Mechanics. Benjamin Cummings. p. 11. ISBN 0131118927.

External links

• Normalization.

• Quantum mechanics