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Since energy can always be shifted by a constant value without changing anything, why do books on quantum mechanics bother carrying the term $\hbar\omega/2$ around?

To be precise, why do we write $H = \hbar\omega(n + \frac{1}{2})$ instead of simply $H = \hbar\omega n$.

Is there any motivation for not immediately dropping the term?

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    $\begingroup$ It is also matter of definition. If you assume from scratch that the Hamiltonian is (up to constants) $H=P^2+X^2$, you must keep the zero point energy $h_0$, since it is the bottom of the spectrum of $H$. Otherwise you should say that the Hamiltonian the harmonic oscillator is $H-h_0I$. The physical problem is whether or not there are physical ways, in quantum mechanics, to distinguish between the two choices. $\endgroup$ – Valter Moretti Mar 5 '14 at 14:30
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    $\begingroup$ The zero-point energy is the difference of ground state energy and infimum of the potential. Even if you shift energy zero, you'll still have classically possible energies, which are below ground quantum state energy. This is the essence of zero-point energy that quantum system never goes lower although classically it could. $\endgroup$ – Ruslan Mar 5 '14 at 16:43
  • $\begingroup$ @ValterMoretti : You pose an interesting question in your last sentence. Do you have an answer or a hint where to find a good answer to it? $\endgroup$ – ungerade Oct 15 '18 at 14:03
  • $\begingroup$ Actually NOT :) ... $\endgroup$ – Valter Moretti Oct 15 '18 at 17:58
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Consider a potential, which approximately can be described by two harmonic oscillators with different base frequencies, for example (working in dimensionless units) $$U=1-e^{-(x-4)^2}-e^{-\left(\frac{x+4}2\right)^2}$$

It will look like

enter image description here

Now let's look at two lowest energy states of the Hamiltonian

$$H=-\frac1m \frac{\partial^2}{\partial x^2}+U,$$

taking for definiteness $m=50$, so that the lowest energy states are sufficiently deep. Now, at the origin of oscillator at left it can be shown to be

$$U_L=\frac14(x+4)^2+O((x+4)^4),$$ and the for right one we'll have $$U_R=(x-4)^2+O((x-4)^4)$$

If two lowest levels are sufficiently deep that their wavefunction don't overlap, then we can approximate them as eigenstates of each of the harmonic oscillators $U_L$ and $U_R$. See how these two states look:

enter image description here

You wanted to remove zero of the total energy by shifting the potential. Of course, you could do this for a single oscillator. But now you have to select, which one to use. And if you select some, you'll still get zero-point energy for another.

Thus, this trick isn't really useful. It tries to just hide an essential feature of quantum harmonic oscillator and quantum states in general: in bound states there is lowest bound on energy, which can't be overcome by the quantum system, although classically the energy could be lower.

Zero-point energy is the difference between minimum total energy and infimum of potential energy. It can't be "dropped" by shifting the potential energy.

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    $\begingroup$ I wouldn't say that the trick is not useful. In many areas it is very useful to drop the term, and in some (QFT) it is even necessary to remove an infinity of zero-point energies. But you have a good point there! If the harmonic oscillator is just an approximation to a more complicated potential it does indeed matter where you have your zero-point energy. This is in my opinion the most satisfactory answer so far. $\endgroup$ – Friedrich Mar 6 '14 at 19:25
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It depends what you're doing, and indeed most of the quantum optics literature dismisses the term as it does not contribute to the dynamics. However, it is important that beginning students form an intuition for how and where zero-point energies come in, and why they are necessary.

Take a look at the eigenfunctions of the harmonic oscillator, in position space:

harmonic oscillator wavefunctions

Notice, in particular, the behaviour at the classical turning points, where the baselines cross the potential. These are the inflection points of the wavefunctions, where the oscillatory behaviour turns into exponential decay. Even for the ground state, these two points must be spatially separated, to allow the exponential decay on the left to turn round into a decreasing function and match into exponential decay on the right, and for these two points to be separated the energy of the ground state needs to be separated from the bottom of the well. This is the essence of the zero-point energy, and until you internalize all the implications of 'classically allowed' and 'classically forbidden' on the wavefunction, it's best to be explicitly reminded that it exists.

On the other hand, once you've done that, there is little point in lugging that term around. If you dig a little deeper into the literature, you'll see people start to drop the term in settings where it is not important. Some examples:

and many, many others. For a good look at what people actually use in the literature, I would recommend searching for 'quantum harmonic oscillator' on the arXiv. This will turn up many papers you won't understand, but it is not that complicated to discard the ones that don't have QHO hamiltonians in them, and distinguish the ones that use hamiltonians of the form $\tfrac1{2m}p^2+\tfrac12 m\omega^2 x^2$ from the ones that use the form $\hbar\omega a^\dagger a$.

It's also worth mentioning that you can't always drop the term. In quantum field theory in particular, you are often faced with a system that is an infinite collection of harmonic oscillators, for which vacuum energy must be treated carefully. On another branch of that, zero-point energies can have measurable effects, for example through the Casimir effect, in which case you obviously can't neglect it.

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  • $\begingroup$ This is a nice graph and I did't know about those inflection points. But I don't need zero point energy for any of that. In quantum field theory you usually drop the term and the Casimir effect can be explained without it: arxiv.org/abs/hep-th/0503158 $\endgroup$ – Friedrich Mar 5 '14 at 15:13
  • $\begingroup$ I'm not sure what you mean by not needing zero point energy for that. Given any potential, the ground state energy will be shifted up from the bottom of the well because of this, and you don't know by how much until you calculate. You can call it something else, but you cannot get away without it. $\endgroup$ – Emilio Pisanty Mar 5 '14 at 16:07
  • $\begingroup$ Regarding the Casimir effect, there are indeed good explanations for the force that do not explicitly invoke it zero-point energy. However, it is a good canonical example of the general fact that it is very naïve to expect zero-point energies to never have any effect or introduce any complications. It might benefit you to keep an open mind at this stage. $\endgroup$ – Emilio Pisanty Mar 5 '14 at 16:09
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    $\begingroup$ You don't need zero-point energy because you can just shift the whole picture down by hw/2. Nothing changes. When talking about wavefunctions in position space, I agree, it is simpler to keep the term. But one usually passes to a treatment with creation and annihilation operators pretty quickly and I do not see why the term is dragged along. $\endgroup$ – Friedrich Mar 5 '14 at 20:16
  • $\begingroup$ That may be, and you're entitled to your opinion. The explanation to your question is there, but if you don't want to see it then it's also fine. Good day! $\endgroup$ – Emilio Pisanty Mar 5 '14 at 20:59
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I disagree that :

Since energy can always be shifted by a constant value without changing anything,

You are maybe thinking of classical potential energy , but the mass of a proton is fixed, for example, it cannot be shifted by a constant value, and at rest $E=mc^2$. This statement is not general and can only be true for the solutions of non relativistic equations.

Edit after comments:

After the comments I realized the question is about a change of the zero of energy, that would not affect the energy levels but the y axis of the potential, which would acquire a negative lower point, so that the first energy level is at 0.

harmosc

This change would only introduce an overall phase factor ( see answer by dextercioby) in the time dependent solutions.

The harmonic oscillator is a very useful quantum mechanical solution because all symmetric potentials have as a first term in their series expansion the x**2. Thus it is extensively used in most many body problems in chemistry, and not only, to model the different collective potentials arising in lattices.

The reason then is for simplicity and esthetics, not to introduce extra complexity in the form of the potential so that the generic equation is described by the simplest functional form of the potential, x**2.

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    $\begingroup$ Anna, if one shifts the ground state energy, one also shifts all other energy levels in the same way. So $\frac{1}{2}\hbar\omega\rightarrow 0$ and $\frac{3}{2}\hbar\omega\rightarrow\hbar\omega$, etc. Also, the binding energy wouldn't be affected because the energy of the transition from bound to non-bound (usually set to zero) would also be shifted, preserving energy differences. (This is all non-relativistic.) $\endgroup$ – BMS Mar 5 '14 at 15:26
  • $\begingroup$ I did't mean you could change any energy at will. I was talking about the total energy (Hamiltonian) of a system. A change of which does not have an influence, as BMS elaborated. $\endgroup$ – Friedrich Mar 5 '14 at 15:47
  • $\begingroup$ @BMS I do not think it will be an exact solution of the equation if you do that, i.e. be in the wavefunction . After all they did not put that 1/2 arbitrarily there, it comes from the solution. hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html#c1 $\endgroup$ – anna v Mar 5 '14 at 17:14
  • $\begingroup$ @Friedrich The formula comes when one operates with the energy operator on the wavefucntions that are the solutions of the shcroedinger equation for the harmonic potential. it is not arbitrary. $\endgroup$ – anna v Mar 5 '14 at 17:34
  • $\begingroup$ Good point anna. Though the energy operator (hamiltonian) itself can be shifted by some constant. Classically, this is just like setting the potential energy at some arbitrary position to be whatever you like. So in QM the $\frac{1}{2}\hbar\omega$ comes from using the classical choice for zero potential energy. One could instead choose for the hamiltonian $H=p^2/2m + kx^2/2 - \hbar\omega/2$. With this choice, I believe $H\psi_{GS}$ would yield $(\hbar\omega/2-\hbar\omega/2)\psi_{GS}$ $\endgroup$ – BMS Mar 5 '14 at 18:06

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