Does quantum tunneling itself result in the collapse of the quantum object's wave function? So, as a hypothetical scenario, suppose you have a two-slit experiment, but instead of two slits, you have two slit-sized barriers that photons have some small probability of tunneling across. Would you get the interefence bands that you get with two slits, or would you get one band for each of the two barriers? Getting interference bands would mean, I believe, that tunneling does not itself result in the collapse of the wave function.
Quantum Mechanics – Does Quantum Tunneling Result in the Collapse of the Wave Function?
quantum mechanicsquantum-tunnelingwavefunction-collapse
Related Solutions
The other answers here, while technically correct, might not be presented at a level appropriate to your apparent background.
When the electron interacts with any other system in such a way that the other system's behavior depends on the electron's (e.g., it records one thing if the electron went left and another if it went right), then the electron no longer has a wave function of its own: the electron+"detector" system has a joint state. The two are entangled.
The electron doesn't have to "know" anything. The simple physical interaction results in a state vector which, by the laws of quantum mechanics, will preclude interference by any of the subsystems of this larger system. That said, the joint state can itself show a kind of "interference effect" (though not the kind you normally think of in the two-slit experiment).
If this entanglement is well-controlled (as in a lab), then (a) showing this "joint interference" might be practical, and (b) undoing the entanglement is also possible, thus restoring the electron's sole superposition. This is how we know that it hasn't "collapsed."
But if the entanglement is caused by stray photons, air molecules, etc., then any hope of controlling them becomes almost immediately dashed, and we can no longer exhibit interference in practice. From here on out, the system will appear to behave classically, with the different branches evolving independently. This fact is called decoherence. The superposition still hasn't "collapsed," but we no longer have the ability to show or exploit the superposition.
You may notice that this still leaves open a crucial question: when do the many branches become one? This is called the measurement problem, and physicists don't agree on the answer even today.
I think this question arises from a simple misunderstanding of what a wave function is. The wave function of a particle doesn't need to be "wavy". The description of a system in quantum mechanics is always given via its state-vector in the Hilbert space and that can always be translated to the wave function of the said system in a basis of your choice, e.g., the position basis or the momentum basis.
A wave function $\psi(x)$ of a particle in position basis simply gives you the probability amplitude of the particle at position $x$ which is a complex number, i.e., it gives you two bits of information:
- The magnitude gives you the probability (density) that you would find the particle in the vicinity of $x$ if you measure its position.
- The phase gives you the information that you'd need on top of the probability (density) to construct the wave function in some other basis, e.g., the momentum basis, so that you can calculate the probabilities (probability densities) associated with the measurement of its momentum.
So, the point is that there is always a wave function of a particle -- regardless of whether it is very localized and point-like or not.
As to why wave functions are nonetheless called wave functions, I think it's a historic relic. There are two tangible historic reasons that resulted in this naming, I think:
- The position-basis wave function of a particle that has a definite momentum is $\sim e^{ipx}$ and it is actually wavy. These are the famous de Broglie matter waves.
- The time-evolution equation that all wave functions satisfy is called the Schrodinger wave-equation (because it was the equation that was followed by the de Broglie waves, I suppose). One should note that the Schrodinger equation is not exactly a wave-equation although it admits wave solutions. It's more like a diffusion equation with an imaginary diffusion coefficient.
Best Answer
Quantum tunneling does not cause wavefunction collapse. You can usually model it as a source of a reflected wave and a transmitted wave with their respective coefficients just as in classic wave theory (optics, acoustics …). This means the transmitted wave can produce interferences.
For your thought experience, you would therefore get interference bands that depend on the transmission coefficient of the two barriers. This will typically distort the interference pattern you would get without these barriers, and lower its intensity due to reflection.
Hope this helps, and tell me if you need more details.