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As I understand, the energy-time uncertainty principle can't be derived from the generalized uncertainty relation. This is because time is a dynamical variable and not an observable in the same sense momentum is.
Every undergraduate QM book I have encountered has given a very rough "proof" of the time-energy uncertainty relation, but not something that is rigorous, or something even remotely close to being rigorous.
So, is there an actual proof for it? If so, could someone please provide me a link to it or even provide me with a proof? Keep in mind that I am not looking a proof using quantum mechanical principles, as comments below pointed out.

EDIT: All the proofs I have found take the generalized uncertainty relation and say "let $Δτ=σ_q/|dq/dt|$", cf. e.g. this Phys.SE post. But this for me does not suffice as a rigorous proof. People give that Δτ a precise meaning, but the relation is proven just by defining Δτ, so I am just looking for a proof(if there is any) that shows that meaning through mathematics. If no better proof exists, so be it. Then I will be happy with just the proof through which we define that quantity. By defining it in this way, there is room for interpretation, and this shows from the multiple meaning that researchers have given to that quantity (all concerning time of course).

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  • $\begingroup$ The mathematicians call it Cauchy-Schwarz, I believe, at least it can be derived from it: physics.stackexchange.com/q/24116. If you take a class on functional analysis, it's kind of considered "introductory material". Having said that, don't mistake physics for mathematics. $\endgroup$
    – CuriousOne
    Commented Jun 2, 2016 at 11:31
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    $\begingroup$ @CuriousOne i know that using the Cauchy-Schwarz inequality one can derive the generalized uncertainty principle, but as time is a dynamical variable(you can't calculate its commutator with the Hamiltonian), i can't see how one can derive the time-energy uncertainty relation from the generalized uncertainty principle. All the books(and links provided) derive it just by saying "let Δτ=σq/|dq/dt|" but this does not satisfy me. $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 11:40
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    $\begingroup$ That's where my comment "don't mistake math for physics" comes in. This relation holds independently of the quantum nature of your functions. It's been used many times over in the WIFI algorithms that are built into your cell phone. You can't derive something from quantum theory that does not require any quantization. $\endgroup$
    – CuriousOne
    Commented Jun 2, 2016 at 11:47
  • $\begingroup$ @CuriousOne but is it not derivable from wave mechanics or maybe Fourier analysis? If i said/implied that it had to be derivable FROM quantum mechanics, then i was wrong to do so. $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 11:52
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    $\begingroup$ @CuriousOne But just defining a quantity that has the units of time as Δτ and then giving our own definition makes it what we say it is? My problem is that this proof does not give it the precise meaning that most people say it has. So, i am asking for a proof that actually shows through the mathematics the meaning that people say it has $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 13:46

3 Answers 3

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The main problem is, as you say, that time is no operator in quantum mechanics. Hence there is no expectation value and no variance, which implies that you need to state what $\Delta t$ is supposed to mean, before you can write something like $\Delta E \Delta t\geq \hbar$ or similar.

Once you define what you mean by $\Delta t$, relations that look similar to uncertainty relations can be derived with all mathematical rigour you want. The definition of $\Delta t$ must of course come from physics.

Mostly of course, people see $\Delta t$ not as an uncertainty but as some sort of duration (see for instance the famous natural line widths, for which I'm sure there exist rigorous derivations). For example, you can ask the following questions:

  • Given a signal of temporal length $t$ (it takes $t$ from "no signal" to "signal has completely arrived"), what is the variance of energy/momentum? This can be mapped to the usual uncertainty principle, because the temporal length is just a spread in position space. It is also related to the so-called Hardy uncertainty principle, which is just the Fourier uncertainty principle in disguise and completely rigorous.

  • If you do an energy measurement, can you relate the duration of the measurement and the energy uncertainty of the measurement? This is highly problematic (see e.g. the review here: The time-energy uncertainty relation. Choosing a model of measurement, you can probably derive rigorous bounds, but I don't think a rigorous bound will actually be helpful, because no measurement model probably captures all of what is possible in experiments.

  • You can ask the same question about preparation time and energy uncertainty (see the review).

  • You can ask: given a state $|\psi\rangle$, how long does it take for a state to evolve into an orthogonal state? It turns out that there is an uncertainty relation between energy (given from the Hamiltonian of the time evolution) and the duration - this is the Mandelstamm-Tamm relation referred to in the other question. This relation can be made rigorous (this paper here might give such a rigorous derivation, but I cannot access it).

  • other ideas (also see the review)...

In other words: You first need to tell me what $\Delta t$ is to mean. Then you have to tell me what $\Delta E$ is supposed to mean (one could argue that this is clear in quantum mechanics). Only then can you meaningfully ask the question of a derivation of an energy-time uncertainty relation. The generalised uncertainty principle does just that, it tells you that the $\Delta$ quantities are variances of operators so you have a well-defined question. The books you are reading seem to only offer physical heuristics of what $\Delta t$ and $\Delta E$ mean in special circumstances - hence a mathematically rigorous derivation is impossible. That's not in itself a problem, though, because heuristics can be very powerful.

I'm all in favour of asking for rigorous proofs where the underlying question can be posed in a rigorous manner, but I doubt that's the case here for a universally valid uncertainty relation, because I doubt that a universally valid definition of $\Delta t$ can be given.

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  • $\begingroup$ I think that this question gets the essence of what i am asking! The fact that in the standard "proof" we are defining Δτ leaves room for interpretations like those that you have presented. Isn't this essentially your answer? That i must tell you exactly what Δτ means in order to try to derive it? $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 13:54
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    $\begingroup$ Yes. And I wanted to convey two more things: (i) in certain cases, we know what $\Delta \tau$ is and thus can derive rigorous bounds such as the Mandelstam-Tamm inequality; (ii) in general, I doubt this is true, i.e. I don't think we can just give one definition that holds for all of quantum mechanics and gives an uncertainty principle always. Such a definition would seem unphysical, because something like "elapsed time" has always some arbitrariness to it (when do I start to measure? When do I stop?). $\endgroup$
    – Martin
    Commented Jun 2, 2016 at 15:27
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This is the case. The uncertainty relationship with energy and time is a matter of Fourier analysis. In fact the relationship $\Delta\omega\Delta t \simeq 1$ was know in classical EM and electrical engineering before quantum physics. The use of Fourier analysis in electrical engineering had much the same uncertainty relationship as the reciprocal relationship between frequency and time.

Classical mechanics has Poisson bracket relationships between momentum and position, and quantum mechanics has an operator replacement of these $$ \{q, p\} = 1~\rightarrow~[q, p] = i\hbar. $$ There are no Poisson brackets in Hamiltonian mechanics between time and energy. In quantum mechanics there is by corollary, using the word informally, no time operator. This leads to some interesting complexities with relativistic quantum mechanics and quantum field theory.

Quantum mechanics is a wave mechanics, and the Fourier analytical basis for the time-energy uncertainty is "good enough" to accept it. The physical basis for energy=time uncertainty is strong enough to accept. We just have some distinguishable situation between space and momentum vs time and energy. In some sense this is a mark that is contrary to the Einsteinian idea.

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  • $\begingroup$ But is there an actual proof? A more rigorous(not the mathematician's kind of rigor)? $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 11:47
  • $\begingroup$ @QuanticMan that sounds like you want documented observations. A proof (in this context) is mathematical by definition. Outside of that, nothing is proven, only theorised and observed to match the theory. $\endgroup$
    – Gusdor
    Commented Jun 2, 2016 at 15:33
  • $\begingroup$ I am talking about mathematical proof $\endgroup$
    – TheQuantumMan
    Commented Jun 2, 2016 at 15:56
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Short answer: position momentum uncertainty exists because their operators do not commute. Likewise the time and energy operators do not commute.

Longer answer: First: Physics is an empirical science so "proof" must be an experiment. Sometimes thought experiments will be allowed because they are useful for building and testing models. Mathematical proofs are not physical proofs.

Synthesis in physics involves building models (usually mathematical models) to explain observed behavior. In QM these intellectual edifices have proceeded quite far.

Within the contexts of those math models for QM, I would note the following:

1) The "state" of a QM system is represented by a "wavefunction", \psi, whose amplitude gives information about the expected outcomes of experiments.

2) Various observations are seen as operators on wavefunctions. For example, position is the operator "x". Interestingly in building the models, momentum has been interpreted as "\partial / \partial \Vec(x)" - i.e. gradient. Why? well because it explains observed experiments. (again - the standard of "truth" or "proof" or derivation in physics)

3) Predicting measurements of observables is a functional acting on the wave-function and the operator. So for instance, the expectation of measuring position is <\psi|x|\psi> (using bra ket notation)

4) Position-momentum uncertainty relation: note the momentum operator, P=\partial/\partial \vec(x) does not commute with the position operator. I.e. Px <> xP in fact the anti-commutator [P,x] \defined Px - xP = 1 ( when one notes that I have used units where h-bar = 1 ) Remember to consider P and x as operators and apply the chain rule for differentiation. This anti-commutation of the operators is, within the constructed models on QM, the source of the uncertainty relation. I.e. not all operators commute.

5) Now consider time and energy. In order to be observed, both must be cast as operators acting on wavefunctions. What are the operators for time and Energy? - simple. The time operator is just multiplication by t. The energy operator is E\ defined \partial / \partial t. --- See how the Energy-time uncertainty relation looks like the position momentum uncertainty relation?

Q.E.D. Well as close as Q.E.D. is meaningful for physics since physics<> math. The real crucible comes in how this prediction from the math model used for QM predicts and agrees with experiment. And it does- very well.

Comments:

A. In this sense, we do not prove E-t uncertainty from x-P uncertainty but rather use the mathematical architecture developed for QM in general.

B. Time being a dynamical variable is and is not a problem. If you want to measure time, you must define an operator to do that. In that sense it is not different than measuring position or spin or charge or ...

C. Time as a "special" dynamical variable can be removed by taking a Langrangian point of view, but that is another discussion that leads down the Feynman path.

D. The E-t uncertainty relation drops out quite naturally from a relativistic approach. When 3-vectors become 4-vectors, you better have t-E uncertainty otherwise Heisenberg will be quite angry. So in that sense one could "derive" E-t uncertainty from x-P uncertainty by requiring good behavior under relativistic considerations, but in my mind, just noting that operators don't commute is a more basic approach. (Clearly a personal intellectual aesthetic judgment, but I think shared by many.)

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