Classical probability density

The probability density function of the $n = 30$ state of the quantum harmonic oscillator. The solid plot represents the quantum mechanical probability density, while the dotted line represents the classical probability density. The dashed vertical lines indicate the classical turning points of the system.

The classical probability density is the probability density function that represents the likelihood of finding a particle in the vicinity of a certain location subject to a potential energy in a classical mechanical system. These probability densities are helpful in gaining insight into the correspondence principle and making connections between the quantum system under study and the classical limit.^[1]^[2]

Mathematical background

Consider the example of a simple harmonic oscillator initially at rest with amplitude $A$ . Suppose that this system were placed inside a light-tight container such that one could only view it using a camera which can only take a snapshot of what's happening inside. Each snapshot has some probability of seeing the oscillator at any possible position $x$ along its trajectory. The classical probability density encapsulates which positions are more likely, which are less likely, the average position of the system, and so on. To derive this function, consider the fact that the positions where the oscillator is most likely to be found are those positoins at which the oscillator spends most of its time. Indeed, the probability of being at a given $x$ -value is proportional to the time spent in the vicinity of that $x$ -value. If the oscillator spends an infinitesimal amount of time $dt$ in the vicinity $dx$ of a given $x$ -value, then the probability density $P (x)$ will be

P(x)\propto dt.

Since the force acting on the oscillator is conservative, the motion will be cyclic with some period which we will call $T$ . Since the probability of the oscillator being at any possible position between $x = - A$ and $x = + A$ must sum to 1, we use the normalization

\int _{-A}^{A}P(x)\,dx=1=N\int _{t_{i}}^{t_{f}}dt,

where $N$ is the normalization constant. Since the ball covers this range of positions in half its period (a full period goes from $- A$ to $+ A$ then back to $- A$ ) the integral over $t$ is equal to $T /2$ , which sets $N$ to be $2/ T$ .

Using the chain rule, we can put $dt$ in terms of the height at which the ball is lingering by noting that $dt = dx /(dx / dt)$ , so our probability density becomes

P(x)={\frac {2}{T}}\,{\frac {dx}{dx/dt}}={\frac {2}{T}}\,{\frac {dx}{v(x)}},

where $v (x)$ is the speed of the oscillator as a function of its position. (Note that because speed is a scalar, $v (x)$ is the same for both half periods.) At this point, all we need to do is provide a function $v (x)$ to obtain $P (x)$ . For systems subject to conservative forces, this is done by relating speed to energy. Since kinetic energy $K$ is $.mw-parser-output .frac{white-space:nowrap}.mw-parser-output .frac .num,.mw-parser-output .frac .den{font-size:80%;line-height:0;vertical-align:super}.mw-parser-output .frac .den{vertical-align:sub}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}1⁄2mv2$ and the total energy $E = K + U$ , where $U (x)$ is the potential energy of the system, we have

v(x)={\sqrt {\frac {2K}{m}}}={\sqrt {{\frac {2}{m}}[E-U(x)]}}.

Plugging this into our expression for $P (x)$ yields

P(x)={\frac {1}{T}}{\sqrt {\frac {2m}{E-U(x)}}}.

Though our starting example was the harmonic oscillator, this formula can be generalized for any one-dimensional physical system. Once the potential energy function has been specified, $P (x)$ is readily obtained for any allowed energy $E$ .

Examples

Simple harmonic oscillator

Starting with the example used in the derivation above, the simple harmonic oscillator has the potential energy function

U(x)={\frac {1}{2}}kx^{2}={\frac {1}{2}}m\omega ^{2}x^{2},

where $k$ is the spring constant of the oscillator and $ω = 2 π / T$ is the natural angular frequency of the oscillator. The total energy of the oscillator is Plugging this into the expression for $P (x)$ yields

P(x)={\sqrt {\frac {1}{\pi }}}{\sqrt {\frac {1}{A^{2}-x^{2}}}}.

References

^ Griffiths, David J. (2005). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 978-0-13-191175-8.
^ Robinett, R. W. (1995). "Quantum and classical probability distributions for position and momentum". American Journal of Physics. 63: 823. doi:10.1119/1.17807.

[1] Griffiths, David J. (2005). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 978-0-13-191175-8.

[2] Robinett, R. W. (1995). "Quantum and classical probability distributions for position and momentum". American Journal of Physics. 63: 823. doi:10.1119/1.17807.

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