# Matter Standing Waves means 0 probability?

de Broglie standing waves are waves that represent the probability of finding an electron somewhere. In an atom, these waves must become standing waves in order to ensure that no destructive interference occurs which would mean that the electron probability would become 0, hence not existing at that particular energy level. However, with standing waves, there must be a time when the entire wave is flat (just for one instant when the anti-nodes swap maxima). Because it is completely flat, that must mean the total sum of probabilities of finding the electron is 0 everywhere for an instant. Thus for an instant, the electron of an atom must not exist? This doesn't sound right so what actually happens?

• the de broglie standing waves are a "handwaving" way of introducing probability solutions , simplified. The atom has a quantum mechanical solution that depends on the atom. The wavefunction complex conjugate square represents the probability. Generally this is not time dependent in the sense you are talking about. I am sure if one makes a timedependent solution there will be no problem like the one you visualize with the "simple" de broglie wave analogue. – anna v Jun 22 '19 at 12:02
• Are you familiar with the expression of waves as a complex exponentials? In classical physics that is merely a convenient short-hand notation, but in quantum mechanics is has a more physical meaning. This question is easily addressed in that form ... for those who have the math. – dmckee --- ex-moderator kitten Jun 22 '19 at 15:16