Negative Energy in Quantum Mechanics: Can Particles Exist?

  • #1
I'm coming here from going through the Dirac delta potential in QM to clear my mind about the case when E < 0, which as a result produces the single bound state solution in that potential. The thing that's vexing my soul(if we have one anyways) is the fact that a particle is taken to be in possession of negative energy. Mathematically speaking, how can it be that a particle has negative energy if both the energy that's coming from the kinetic term and that of the potential are strictly positive? Physically speaking, how can it be that a particle possesses negative energy if it exists in this universe at all? Even if it hasn't got any momentum there's potential. And unless you're analyzing it's motion relative to a reference frame (easiest example would be to pick the ceiling as the point where the gravitational potential is zero therefore an immobile body in the room would end up with negative energy), you can't have a particle of negative energy. Even then, it's because of relativity and not the actual true energy and Introduction to QM by Griffiths(any intro textbook to qm for that matter) doesn't take relativity into account. A particle can't actually POSSESS negative energy and exist realistically at the same time I don't think.

I am truly sorry to discuss such petty matter in this forum but I don't have any professor to go to and talk QM with because I'm still 16 attending high school. This kind man deriving the ddf potential bound state solution to Shroed's on youtube circumvents my problem by setting the ddf to ping to -∞ so that by having E < 0 the bound state just mathematically comes by. I thank anyone in advance for any enlightment they can offer :D
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Answers and Replies

  • #2
if both the energy that's coming from the kinetic term and that of the potential are strictly positive?

Maybe I'm missing something, but are you sure the potential is strictly positive? I didn't watch the entire video, but he draws the potential as ##V(x) = -a \delta(x)##, which is definitely not a strictly positive function. In fact, it's (loosely) negative infinity at x=0.
  • #3
That's not the point I was trying to make though. The potential might as well fly off at -∞ and putting E < 0 just spews out the bound state. My concern was the actual choosing of E < 0. How can it be that that isn't just some relative choice of the reference frame but actually a TISE solution with negative energy?
  • #4
The negative energy of the state is an effect of the arbitrary choice of zero potential.

The energy below the potential is an effect of quantum mechanics: the particle spreads out to a "forbidden area", this always happens in bound states (unless your potential goes to +infinity), and it can lead to tunneling if there is some target potential to tunnel into. You will never see your particle in the forbidden area unless you give it enough energy to get above the potential there.
  • #5
The energy sure might be below the potential at some point and therefore give rise to classically forbidden existence of a particle, an "outside of bounds" particle, but how is it that it has E < 0 if V(x) is the ddf? Is it just a matter of what sign you stick in front of the delta function? I'd be very disappointed if it's just that.
  • #6
With a different sign, you don't get any bound states.

E<0 has no physical meaning in this example, only the energy relative to the potential is relevant. Define your potential to have a large positive value and the bound state has E>0.
  • #7
Yeah, figured that out anyways. Thank you for your time.

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