# I Bohr model, Why do we assume a standing wave?

1. Nov 10, 2016

### kidsasd987

Hi, I wonder why we assume the matter wave of an electron is standing wave. Is there any reason why it has to be standing wave?

Is it because standing wave is the right "wave equation solution" that satisfies integer multiple behaviour of bohr model?

2. Nov 10, 2016

### nashed

Disregarding the fact that the bohr model is wrong, the reason I've heard is that a standing wave would explain why the electron doesn't radiate EM waves and spiral down to the nucleus.

3. Nov 10, 2016

### Staff: Mentor

A standing wave is required so that the solution is single-valued at every point in the orbit.

4. Nov 10, 2016

### kidsasd987

Could you specify further? I understood this way. so spatially, if we have a wave equation solution that is not a standing wave, then it would create a interfered wave when it goes one round(sum of the same wave equation with different phases which we cannot find integer related equation). And this feedback process would not give a single solution and that is absurd.

5. Nov 10, 2016

### Staff: Mentor

You would get zero on average everywhere, and it would not be a stationary state.

6. Nov 10, 2016

### kidsasd987

Ah thanks!

7. Nov 10, 2016

### Staff: Mentor

More precisely, the wave function of a bound electron--that is, an electron that is confined in a bound state--is a standing wave (if we ignore the issues with that term--see below). We don't assume this; we derive it by solving the Schrodinger equation with an appropriate potential energy term describing how the electron is bound (for example, the Coulomb potential of the nucleus in an atom), and looking at the time-independent solutions.

Also, the term "standing wave" might be misleading, because it suggests that the bound electron is confined to, for example, a single "orbit" at a fixed radius around the nucleus in an atom. That is not the case. The wave function describing a stationary state of the bound electron in an atom is distributed in all 3 spatial dimensions, and the "nodes" of the distribution (places where the amplitude is zero) are not equally spaced as the "standing wave" analogy suggests. (So, for example, your images in the OP are not descriptions of an actual electron wave function.)

8. Nov 10, 2016

### kidsasd987

Thank you it helped a lot!

9. Nov 11, 2016

### vanhees71

I also want to add that you mix here models. The Bohr-Sommerfeld model is part of "old quantum theory", and it's outdated for more than 90 years now. It's interesting for historians of science to analyze how groundbreaking new insights in the natural sciences are found, but that's all that's interesting about it nowadays.

The "standing-wave picture" is part of modern non-relativistic quantum theory in its "wave-mechanics formulation" a la Schrödinger, which implies that the stationary (time-independent) states are given by the eigenstates of the Hamilton operator. The Schrödinger equation then tells you that the corresponding eigensolutions of the Hamiltonoperator just depend on time via a phase factor $\exp(-\mathrm{i} E t/\hbar)$, and thus the probabilities, given by the modulus of the wave function squared, are indeed time-independent. That's why the stationary solutions of the Schrödinger equation are precisely the eigenstates of the Hamilton operator of the system and that these states factorize in a time-dependent phase factor (which is irrelevant for the physics content of the wave function, i.e., the probability distribution for finding the particle at a place doesn't depend on it) and a position dependent solution of the time-independent Schrödinger equation, which is just the eigenvalue equation for the Hamiltonian. These are then by definition of course standing waves.

10. Nov 11, 2016

### Khashishi

There's no a priori reason why the wave function has to be stationary. The planetary model assumes electrons revolving around the nucleus. You could imagine wave pulses traveling around the nucleus in a similar fashion. It doesn't matter if the field isn't single valued because each time the wave passes a point is at a different time. Bohr assumed a standing wave because that's what works. The actual scientific method goes both ways. Experiment drives models that explain the results, which ideally predict more results which can be measured by experiments. Some cartoon depictions of the scientific method depict people deriving everything from scratch on a chalkboard and telling the experimentalists what the measure. Usually, it's the other way around.

11. Nov 11, 2016

### Staff: Mentor

Not in general, no. But if we are looking for a stationary state, i.e., a state that does not change with time, yes, it does.

And that model is wrong, so it's irrelevant here.

No, he assumed a standing wave because it happened to give a model that worked better than the classical model of "electrons revolving around the nucleus", since the latter model predicted that atoms would collapse, which is obviously wrong. But Bohr's model does not "work" in the sense of accounting for all the experimental knowledge we now have; that's why we don't use it any more.

12. Nov 11, 2016

### Khashishi

Of course. By "it works" I meant "it works better than a model without stationary states, like the planetary model". My interpretation of the OP question is "why did Bohr assume a standing wave rather than a wave that propagates around the nucleus in a circle?" Some of the responses were begging the question. I hope the OP realizes that the Bohr model is obsolete, and wavefunctions do not look like the pictures above, and the OP was simply asking out of historical curiosity.

13. Nov 11, 2016

### Staff: Mentor

Actually, Bohr didn't use standing waves in his model. He assumed that the orbital angular momentum is quantized. (Actually he assumed the "action-angle integral" of the electron's motion is quantized, IIRC, which is equivalent to the orbital angular momentum in this case.)

De Broglie came up with the standing-wave idea several years later, as a way of deriving Bohr's quantization condition.