3

Here is a picture of the energy states of infinite potential well.

Energy levels

We can see That the first level have a half wavelength which fittes with a full wave of the second level. enter image description here

$$\frac{ \lambda _{1} }{2} = \lambda _{2}$$

$$ \Rightarrow \lambda _{1} =2 \lambda _{2}$$

$$ \Rightarrow \frac{1}{E_{1} } = \frac{2}{E_{2} } $$

$$\Rightarrow E_{2} = 2E_{1} $$

We have used the formula:$$E=h \nu $$

$$\Rightarrow E= \frac{hc}{ \lambda } $$

We started with our conception of classical waves, we used Planck's relation $E=h \nu $ and we have concluded $E_{2} = 2E_{1} $. Which we used to know wrong.

We know that $$E_{n}= n^{2} E_{1}$$

Now we start from here:

$$E_{2}= 4 E_{1}$$

$$ \Rightarrow \frac{1}{ E_{1} } = \frac{4}{ E_{2} } $$

$$ \Rightarrow \lambda _{1} = 4 \lambda _{2} $$

$$\Rightarrow \frac{ \lambda _{1} }{2} =2 \lambda_{2} $$

We can represent the last line with a picture

enter image description here

Here two waves fittes in a half wavelength of the first level.

Now my question is why our concept of classical waves doesn't reconcile with the energy level wave functions? Is there any other difference between classical and quantum wave? What's wrong in my analysis?

Qmechanic
  • 201,751

3 Answers3

6

The quantum wavefunction is not a wave in the classical sense. In particular, it is not a wave obeying $E = \hbar \omega$, as electromagnetic waves do.

The term wavefunction is, in this sense, just a bad naming decision. It is no wave, there is nothing oscillating, and it has no connection whatsoever except the mathematical form to physical waves.

ACuriousMind
  • 124,833
  • Agreed - Correct +1, but we can observe intereference and other 'wave' type effects for particles. – tom Nov 18 '14 at 17:52
  • 3
    I like the first paragraph. I don't agree with much of the second. – BMS Nov 18 '14 at 19:55
  • @ACuriousMind: Who told you that the wave-function is not a wave? YES, it is a wave, (in particular it has absolute value and phase) just we don't know of what nature. I remind you that it produces single-particle interference, as the classical waves. But it also produces 2particle interference, which we obtained only in quantum mechanics. What is the nature of the wave-function is the great non-answered question of the quantum mechanics. – Sofia Nov 19 '14 at 11:10
  • 1
    @Sofia: The wave function is simply the coefficients of any state $\lvert \psi \rangle$ when expanded in the position basis of the Hilbert space of states, if such a basis exists. Nothing more. Nothing less. The interference effects can be purely seen from the way the inner products on the Hilbert space behaves. The whole of "wave mechanics" is completely equivalent to "matrix mechanics" (unitary equivalence of Schrödinger and Heisenberg pictures), where the notion of waves never enters. – ACuriousMind Nov 19 '14 at 11:25
2

Wave equations have a long history in physics, they are usually equations involving second derivativs, the solutions are sinusoidal ( sines and cosines) and have been used to model classical waves, starting from water, sound, pressure waves, and finally light classically, with Maxwell's equations.

When Schrodinger's equation was able to reproduce the Bohr model solutions for the atoms, it was a wave equation, it has sinusoidal solutions.

What is the difference?

The difference is in the postulates used to fit/project the mathematical models to the physical data under consideration, and to predict new behaviors for the object of the study.

The postulates for classical waves are that the amplitudes of the solutions correspond in space and time to the energy carried by the wave. That is what was under study and the mathematics was developed.

For quantum mechanics a main postulate is that the square of the wavefunction represents the probability of finding the particle under consideration at that specific (x,y,z,t). So the sinusoidal variations vary the probability not the mass or energy in space and time.

Probability distributions have the same meaning and are accumulated in the same way classically and quantum mechanically. One has to gather many instances to get the probability distribution experimentally. For throwing dice, it is flat (1,2,3,4,5,6) , for the electron in its orbital around the nucleus it is a sinusoidal function. This has been seen in quantum mechanical interference phenomena, as in the two slit experiment.

Now the case of the photon/light is a bit special as Maxwell's equation is used both classically and quantum mechanically ( by turning it into an operator form) and thus the continuity between classical and quantum mechanical formulations at their interface is retained. The photon in the double slit experiment displays the interference a single photon at a time, displaying the probability aspect, and the electromagnetic wave has the frequency that will display the same interference.

anna v
  • 233,453
1

Comment to the question (v2): The question formulation conflates on one hand a non-relativistic particle in an infinite potential well/box, which has a quadratic dispersion relation $E \propto p^2$; and on another hand a massless relativistic dispersion relation $E \propto p$.

See this Phys.SE question and links therein for a similar misunderstanding.

Qmechanic
  • 201,751