# Momentum-Position vs. Energy-Time Uncertainty Relations

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## Summary:

I have read that the uncertainty relation for momentum and position is unrelated to the uncertainty relation for energy and time. But I am confused because when I think about it, the two uncertainty principles actually seem like two sides of the same coin. The "uncertainty in position" can instead be thought of as the spatial delocalization of the wave function, in analogy with the "lifetime" in the energy-time uncertainty relation, which is the temporal delocalization of the wave function.

## Main Question or Discussion Point

In a few textbooks in introductory quantum mechanics which I have looked through (e.g. Griffiths), it is heavily emphasized that the momentum-position uncertainty relation has a completely different meaning from the energy-time uncertainty relation, and that they are quite unrelated and only superficially look similar. But I don't see why this is the case. Please let me explain.

From what I know, there are two ways to think about the momentum-position uncertainty principle. One way (which I believe is only valid in non-relativistic quantum mechanics) is that position should be treated as a dynamical variable that is an attribute of the particle. Then, the momentum-position uncertainty principle says that if the probability distribution in the outcome of a position measurement has a standard deviation Δx, then the probability distribution in the outcome of a momentum measurement must have a standard deviation no less than ħ/(2Δx).

But there is another way to think about the momentum-position uncertainty principle. We can treat the position as a parameter similar to time, rather than as an attribute of the particle. Then, the momentum-position uncertainty principle states that if the probability density of detecting a particle at a certain spatiotemporal location has a spatial dependence following a distribution with a standard deviation Δx, then the probability distribution in the outcome of a momentum measurement at that time must have a standard deviation no less than ħ/(2Δx).

As far as I can tell, these two forms of the momentum-position uncertainty principle are equivalent (within the context of non-relativistic quantum mechanics) since if the probability distribution in the outcome of a position measurement has a standard deviation Δx, that precisely means that if one places a particle detector in space, then the probability density that the detector will detect the particle will vary with the position of the detector according to a distribution with standard deviation Δx.

But the second of the above two interpretations of the momentum-position uncertainty principle (which treats position as a parameter) seems strikingly similar in meaning to the energy-time uncertainty principle. The energy-time uncertainty principle states that if the probability density of detecting a particle at a certain spatiotemporal location has a temporal dependence following a distribution with a standard deviation Δt, then the probability distribution in the outcome of an energy measurement must have a standard deviation no less than ħ/(2Δt). Compare this with the momentum-position uncertainty principle, which (as already mentioned) states that if the probability density of detecting a particle at a certain spatiotemporal location has a spatial dependence following a distribution with a standard deviation Δx, then the probability distribution in the outcome of a momentum measurement must have a standard deviation no less than ħ/(2Δx).

It seems that the momentum-position uncertainty principle is indeed totally analogous to the energy-time uncertainty principle since the former has two equivalent formulations (one with position as a dynamical variable, and the other with position as a parameter). Even though time can only be treated as a parameter, the fact that we can always change from treating position as a dynamical variable to treating position as a parameter seems to mean that the two uncertainty principles are actually intimately related, one focusing on the spatial dependence of the wave function and the other focusing on the temporal dependence of the wave function.

But then, what is the point that these textbooks (such as Griffiths) seem to make? Is there any mistake in my argument above? Any clarification is appreciated.

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PeroK
Homework Helper
Gold Member
But then, what is the point that these textbooks (such as Griffiths) seem to make? Is there any mistake in my argument above? Any clarification is appreciated.
I want to point out that Griffiths also says on page 114 that in a relativistic theory you would expect the energy-time relation as a "necessary concomitant" to the position-momentum relation.

Griffiths is presenting a non-relativistic theory of QM and he avoids using relativistic arguments. After all, once you have started with the fundamentally non-relativistic SDE, it's not really valid to bring in relativistic arguments here and there.

BvU
hilbert2
Gold Member
You don't need the time dependent Schrödinger equation at all to show that the p-x-uncertainty relation holds. It follows from the commutation relation of $\hat{p}$ and $\hat{x}$.

The energy-time uncertainty follows from the TDSE and practically says that if a quantum state is a mixture of many energy eigenstates, then the position and momentum space wavefunctions change quickly so that the probability densities become something else in a short time. In the opposite situation, the probability density $|\psi (x)|^2$ of a one-particle wavefunction remains constant if $\psi$ is an energy eigenstate.

BvU
Demystifier
In your interpretation of time uncertainty, you need something like probability density $\rho(t)$ or $\rho(x,t)$ as a function of time or spacetime. This probability density has to satisfy a normalization condition
$$\int_{-\infty}^{\infty} dt \, \rho(t) =1$$
or
$$\int_{-\infty}^{\infty} dt \int d^3x \, \rho(x,t) =1$$
In quantum mechanics, the probability density is expected to be something like
$$\rho(x,t) =|\psi(x,t)|^2$$
so the normalization condition should be something like
$$\int_{-\infty}^{\infty} dt \int d^3x \, |\psi(x,t)|^2 =1$$
All this seems innocent, but here is the problem. When $\psi(x,t)$ is a solution of the Schrodinger equation, then in fact
$$\int d^3x \, |\psi(x,t)|^2 ={\rm constant}$$
so
$$\int_{-\infty}^{\infty} dt \int d^3x \, |\psi(x,t)|^2 ={\rm constant}\int_{-\infty}^{\infty} dt =\infty$$
Hence the wannabe probability density $\rho(x,t) =|\psi(x,t)|^2$ cannot be properly normalized. A possible way out is to consider a finite region of time which gives
$$\int_{t_1}^{t_2} dt ={\rm finite}$$
but most physicists don't find it very appealing.

hilbert2
Gold Member
You can also define an autocorrelation function $C(t) = \langle \psi(0)|\psi (t)\rangle$ and follow how quickly it approaches 0 when time passes.

Momentum can change while Energy stays the same.

PeterDonis
Mentor
2019 Award
Momentum can change while Energy stays the same.
How does this relate to the topic of this thread?

Demystifier
Momentum can change while Energy stays the same.
It's completely irrelevant but the opposite is also possible, e.g. when you fall down with a parachute.

vanhees71
Gold Member
2019 Award
The time-energy uncertainty relation is indeed different in nature from the usual uncertainty relation concerning observables since in QT time is not an observable at all. For the same reason it's such a conceptually complicated issue to precisely define the "tunnel time" of particles through a potential barrier.

Is there some sort of symmetry between Et and px? What would be the conserved quantity?

vanhees71