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7.5 Quantum-Classical Correspondence

One of the interesting questions raised by the fact that we can solve both the quantum and the classical problem exactly for the harmonic oscillator, is ``Can we compare the Classical and Quantum Solutions?''

  
Figure 7.2: The correspondence between quantum and classical probabilities
\includegraphics[width=7cm]{Figures/QuantClass.ps}

In order to do that we have to construct a probability for the classical solution. The variable over which we must average to get such a distribution must be time, the only one that remains in the solution. For simplicity look at a cosine solution, a sum of sine and cosines behaves exactly the same (check!). We thus have, classically,

x = Acos($\displaystyle \omega$t),        v = - A$\displaystyle \omega$sin($\displaystyle \omega$t). (7.29)

If we substitute this in the energy expression, E = $ {\frac{1}{2}}$mv2 + $ {\frac{1}{2}}$m$ \omega^{2}_{}$x2, we find that the energy depends on the amplitude A and $ \omega$,

E(A) = $ {\frac{1}{2}}$mA2$\displaystyle \omega^{2}_{}$sin2($\displaystyle \omega$t) + $ {\frac{1}{2}}$m$\displaystyle \omega^{2}_{}$A2cos2($\displaystyle \omega$t) = $ {\frac{1}{2}}$m$\displaystyle \omega^{2}_{}$A2 (7.30)

Now the probability to find the particle at position x, where - A < x < A is proportional to the time spent in an area dx around x. The time spent in its turn is inversely proportional to the velocity v

$\displaystyle \mathcal {P}$class(x)dx $\displaystyle \propto$ $\displaystyle {\frac{1}{\vert v(x)\vert}}$dx (7.31)

Solving v in terms of x we find

| v(x)| = $\displaystyle \omega$$\displaystyle \sqrt{A^{2}-x^{2}}$ (7.32)

Doing the integration of 1/v(x) over x from - A to A we find that the normalised probability is

$\displaystyle \mathcal {P}$class(x)dx = $\displaystyle {\frac{1}{\pi\sqrt{A^{2}-x^{2}}}}$dx. (7.33)

We now would like to compare this to the quantum solution. In order to do that we should consider the probabilities at the same energy,

$ {\frac{1}{2}}$m$\displaystyle \omega^{2}_{}$A2 = $\displaystyle \hbar$$\displaystyle \omega$(n + $ {\frac{1}{2}}$), (7.34)

which tells us what A to use for each n,

A(n) = $\displaystyle \sqrt{\frac{\hbar}{m \omega} (2n+1)}$. (7.35)

So let us look at an example for n = 10. Suppose we choose m and $ \omega$ such that $ \sqrt{\frac{\hbar}{m \omega}}$ = 10-10 m. We then get the results shown in Fig. 7.2, where we see the correspondence between the two functions.


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Next: 8. The formalism underlying Up: 7. The Harmonic oscillator Previous: 7.4 A few solutions

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk