Quantum Harmonic Oscillator: Negative Kinetic Energy Question.

[BAnders1]

Hello everybody,

I noticed these questions are lengthy. If you want to skip my introduction, just scroll down to the questions. I put *** next to each one.

I just started Quantum Theory I this semester and I have a question (actually two questions) regarding the quantum harmonic oscillator (QHO). (By the way, we are only discussing one dimension here).

In the QHO (more specifically in the ground state of the QHO), we find that there is a probability that the particle in question may be located beyond the "classical turning points," that is, the particle's potential energy is greater than its total energy, meaning it has a negative kinetic energy. This is weird.

After coping with this, I figured that the particle's velocity must be an imaginary number (because I'm not willing to cope with a negative mass just yet). Now without proof I was told that "the outcome of a measurement has got to be real" (David J Griffiths - Intro to Quatum Mechanics), and I don't really doubt this (though I've gotten some pretty weird measurements in my labs).

***My first question is: Since an imaginary velocity means an imaginary momentum, does this mean that I cannot measure the momentum of a QHO particle if I already measured its position to be beyond the classical turning points? Now I'm aware that after I measure the position, the wave function collapses and I can't measure the momentum, but what I mean is...

If there is (let's say) a 16% chance that the particle is located beyond the classical turning points (so that its momentum is imaginary), does that mean that there is a 16% chance that when I measure its momentum, I find that it does not have one (or something along these lines).

***Now you HAVE to read my second question's intro because I find it necessary that you should know where this question is coming from (sorry for tricking you). I determined that there is about a 15.8% chance that a particle in a QHO has a negative kinetic energy. I determined that this probability does not depend on the particle's mass. In fact, it doesn't depend on anything! (I determined that the probability is that of a normal distribution from z= -1/sqrt(2) to 1/sqrt(2), if that means anything to you.) 15.8% of the time an oscillating particle has an imaginary velocity. This can't be right; when I flick my car's antenna, I don't see any quantum effects.

 

[Nugatory]

In the QHO (more specifically in the ground state of the QHO), we find that there is a probability that the particle in question may be located beyond the "classical turning points," that is, the particle's potential energy is greater than its total energy, meaning it has a negative kinetic energy. This is weird.

Careful... you measured the position, not the energy, so you have no idea what the energy is. You are implicitly assuming that both can be measured simultaneously, and of course they cannot.

Whatever you used to measure the position had to interact with the particle. That interaction had to exchange some energy with the particle; when you find the particle in the classically forbidden region beyond the turning point the interaction will have supplied the necessary additional energy. The energy of the total system (particle, oscillator, measuring device) is conserved, but the particle can end up with more or less energy.

After coping with this, I figured that the particle's velocity must be an imaginary number

You haven't measured it, so you have no idea what it is, but if you do measure it you will get a real number consistent with a real kinetic energy. The total energy will be greater than the initial oscillator energy if a previous position measurement had found the particle in the forbidden region and added some energy in the process.

My first question is: Since an imaginary velocity means an imaginary momentum...

No, because there is no imaginary velocity. The kinetic energy will always be non-negative.