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I understand that we want to solve for non-zero values of wave function. I always thought that is to avoid the obvious answer to Schrodinger equation. But from physical standpoint, if we have a particle of mass $m$, is it really impossible for it to have energy of zero? From mathematical point of view shouldn't the ground state energy of every system be zero? If yes, what does that mean? Nothingness. Void as ground state?

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    $\begingroup$ But isn't there a zero-energy ground state for the particle-in-the-box, by taking $n=0$ in $E_n=\frac{n^2\hbar^2\pi^2}{2ma^2}$? $\endgroup$ – probably_someone Jun 21 '17 at 4:36
  • $\begingroup$ Did you mean the quantum harmonic oscillator? $\endgroup$ – Avantgarde Jun 21 '17 at 5:08
  • $\begingroup$ @probably_someone exactly that was my question. A textbook I read says that the ground state is at n=1. I do not understand whether avoiding zero is arbitrary or theoretically/physically meaningful. $\endgroup$ – Kinformationist Jun 21 '17 at 5:13
  • $\begingroup$ @probably_someone n=0 is not a legitimate choice for the quantum number in this case, for example if you put that into your wavefunction for the particle in a box you obtain nonsense $\endgroup$ – R. Rankin Jun 21 '17 at 5:27
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    $\begingroup$ Another way to formulate the issue is that if we allowed a ground state with $n=0$, it wouldn't be normalizable, as the wave function is identically zero. If we would've allowed $n=0$, and supposedly prepare a system in this ground (stationary) state we would have $\int \psi \psi^*=0$, instead of $\int \psi \psi^*=1$ as required from a quantum state. $\endgroup$ – Yair M Jun 21 '17 at 7:56
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As far as most textbooks on (non-relativistic) quantum mechanics go, we do not consider the solution for $n=0$ because it gives us a trivial solution (and we interpret it to mean that there is no particle inside the box/well).

However, if there were a ground state with zero energy for a square well potential, it would imply that (since the particle has zero energy), it will be at rest inside the square well, and this will clearly violate Heisenberg's uncertainty principle!

By confining a particle to a very small region in space, it acquires a small but finite momentum. So, if the particle is restricted to move in a region of width $\Delta x \sim a$ (i.e., the entire length of the well), we can calculate the minimum uncertainty in momentum (using the uncertainty principle) and it comes out to be $\Delta p \sim \hbar/a$. And, this in turn gives us the minimum kinetic energy of the order $\hbar^2/(2ma^2)$. This (qualitatively) agrees with the exact value of the ground state energy.

So physically, the existence of a zero-point energy is a necessary feature of a quantum mechanical system. It indicates that the particle should exhibit 'a minimum motion' due to localization. Classically, the lowest possible energy of system corresponds to the minimum value of the potential energy (with kinetic energy being zero). But in quantum mechanics,the lowest energy state corresponds to the minimum value of the sum of both potential and kinetic energy, and this leads to a finite ground state or zero point energy.

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  • $\begingroup$ I understand your line of reasoning. Shouldn't this be mathematically addressed in the very formulation of Schrodinger equation? Is it fine to put an extra statement saying let's avoid some mathematical solution to keep it consistent with uncertainty principle? $\endgroup$ – Kinformationist Jun 21 '17 at 16:15
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    $\begingroup$ Well, Heisenberg's uncertainty principle follows from the postulates of quantum mechanics and the Schrödinger equation or the time evolution of the system is itself another postulate. So, as far as I understand it, everything is consistent with the postulates of quantum mechanics here, as it should be. We are not adding "extra statements" arbitrarily, if that's what you are asking. $\endgroup$ – noir1993 Jun 21 '17 at 16:25
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The zero of the energy is completely arbitrary, as the zero of time or space.

In fact, suppose that the ground state energy of $H $ is $a $, then $H-a I$, where $I $ is the identity operator, has ground state energy zero and the same eigenvectors of $H $. In addition, it generates the same time evolution (apart from a non-physical phase factor). Therefore it is, from the physical standpoint, indistinguishable from the original one.

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  • $\begingroup$ What you are suggesting essentially amounts to perturbing the Hamiltonian, right? Basically, perturbations of order greater than one will vanish for a constant perturbation in general. We can calculate the wave function as well. But is this true always? We will run into trouble if the (unperturbed) energy spectrum is degenerate (e.g., while solving for a 3D box). $\endgroup$ – noir1993 Jun 21 '17 at 7:04
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    $\begingroup$ We can shift a Hamiltonian by any constant, but the problem here is that, because you'd choose an $\hbar$-dependent constant, you wouldn't be quantising the classical Hamiltonian, which is written on the assumption $\hbar=0$. So there actually is a physical interpretation to be made of the zero-point energy that this just-shift-the-potential argument overlooks. noir1993 has provided the box's interpretation; for the harmonic oscillator it's that $\left[ a,\,a^\dagger\right]\propto\hbar$ because $\left[ x,\,p\right]\propto\hbar$. $\endgroup$ – J.G. Jun 21 '17 at 7:21
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    $\begingroup$ @noir1993 If we can't analytically deduce the new Hamiltonian's spectrum we need small perturbations to use Taylor series, but in this case the spectrum is trivial to obtain. $\endgroup$ – J.G. Jun 21 '17 at 7:23
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    $\begingroup$ @J.G. That an Hamiltonian is the quantization of a classical symbol is a matter of convenience, not a physical requirement. The physical requirement is that in the limit you will obtain the correct classical behavior. And that is still true if the Hamiltonian is shifted by the $\hslash$-dependent zero point energy. In addition, as far as I know, only energy differences are observable experimentally (e.g. measuring the frequency of the emitted or absorbed radiation); therefore the ground state energy is not measurable. $\endgroup$ – yuggib Jun 21 '17 at 8:09
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    $\begingroup$ This answer is somewhat off the track: while you can shift total energy by an arbitrary amount, you can't avoid the nonzero mean kinetic energy in the ground state. $\endgroup$ – Ruslan Jun 21 '17 at 10:33
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My simple version of the answer:

  • because for the energy to be zero, time must be stopped

I'm sorry I don't know how to state that in the maths you're looking for, perhaps someone else can, I suspect that's what noir1993 has provided.

I think in terms you can visualise, it's kind of like saying that yes it's "at rest" from a perspective of Newtonian physics, but nothing is "at rest" from the quantum perspective unless time actually stops.

I don't know if that helps, but I was just fascinated by the question and wanted to chime in ( yes I know this isn't facebook )

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