Quantities in the Heisenberg Uncertainly Principle

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SUMMARY

The discussion centers on the Heisenberg Uncertainty Principle (HUP) and its application to different physical quantities. It clarifies that energy and position are not conjugate observables, while energy and time are, leading to different uncertainty relations. The conversation highlights that time is a parameter in quantum mechanics, not an observable, which complicates the energy-time uncertainty relation. The mathematical formulation of the HUP is presented, emphasizing the distinction between observables and parameters in quantum theory.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the Heisenberg Uncertainty Principle.
  • Familiarity with the concept of conjugate observables in quantum theory.
  • Knowledge of mathematical operators and their role in quantum mechanics.
  • Basic grasp of Hamiltonian mechanics and energy eigenstates.
NEXT STEPS
  • Study the mathematical derivation of the Heisenberg Uncertainty Principle in quantum mechanics.
  • Explore the concept of conjugate observables and their implications in quantum systems.
  • Research the role of time as a parameter versus an observable in quantum mechanics.
  • Examine the energy-time uncertainty relation and its applications in quantum field theory.
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Physics students, quantum mechanics researchers, and anyone interested in the foundational principles of quantum theory and the implications of the Heisenberg Uncertainty Principle.

Kremmellin
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I'd like to ask a question about quantities in the Heisenberg Uncertainly Principle
Hello, I am a Brazilian Physics student and would like to ask a question. Why are not all physical quantities related to each other by the degree of precision in the Heisenberg Uncertainty Principle? For example, why is it possible to determine the energy and position of a particle without its precision being inversely proportional to each other, but it is not possible to calculate energy and time with the same precision?
 
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Hi, i am a Brazilian Physics student too.

When do you mean why, what exactly do you are waiting for an answer?
I mean, i believe there is two ways to answer it

Mathematically, which will show to you why this happens in some cases, and why it does not happen in other cases. And that is it, "it is true because we can show that it is true, and we observe in nature that it is true".
Or answering using an physical interpretation (maybe philosophically) to try to explain why make senses this is true, and not why this is true.
 
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Kremmellin said:
For example, why is it possible to determine the energy and position of a particle without its precision being inversely proportional to each other, but it is not possible to calculate energy and time with the same precision?
Because energy and position are not conjugate variables, but energy and time are:
https://en.wikipedia.org/wiki/Conjugate_variables
 
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lomidrevo said:
energy and position are not conjugate variables, but energy and time are

Actually, the correct term is conjugate observables (not "variables"). And time is actually not an observable, so there actually is not an energy-time uncertainty principle in the same way that there is, for example, a momentum-position uncertainty principle. There is something that is similar to an uncertainty principle with energy and time, but it takes more work to derive it. See, for example, here:

https://math.ucr.edu/home/baez/uncertainty.html

(Note that Baez in that article is using "operator" to mean the same thing as I used "observable" to mean above.)
 
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Kremmellin said:
Summary:: I'd like to ask a question about quantities in the Heisenberg Uncertainly Principle

Hello, I am a Brazilian Physics student and would like to ask a question. Why are not all physical quantities related to each other by the degree of precision in the Heisenberg Uncertainty Principle? For example, why is it possible to determine the energy and position of a particle without its precision being inversely proportional to each other, but it is not possible to calculate energy and time with the same precision?
Let's take the example of a particle in an infinite square well. And consider an energy eigenstate. A measurement of energy has no uncertainty. And, as E=p22m, then a measurement of the magnitude of the momentum has no uncertainty. All the uncertainty is in the direction of the momentum: we know the magnitude of a measurement of momentum, but there's an equal probability of its being positive or negative. That means that the expected value of momentum is actually zero, and there is a uncertainty (variance or standard deviation) in the outcomes of momentum measurements.

The question you should be asking is why is the energy-time uncertainty relation similar to the HUP?

Note that time in QM is a parameter and not an observable. So, you should also be asking what does "precision in a measurement of time" mean?

Also, if you have an energy eigenstate, where the uncertainty in energy is zero, then the uncertainty in time is infinite! What is that supposed to mean? Does that mean that if you measure energy in an energy eigenstate, then a "measurement" of time is impossible?
 
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The general Heisenberg uncertainty principle (HUP) for observables (and note that time is NOT an observable in QT for the reason that energy must be bounded from below to have a stable ground state; the energy-time uncertainty relation(s) have NOT the same meaning as the HUP for observables!):
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
Then for ##\hat{B}=\hat{H}## note that
$$[\hat{A},\hat{H}]=\mathrm{i} \hbar \mathring{\hat{A}},$$
where ##\mathring{\hat{A}}## is the operator representing the time derivative of ##\hat{A}## (assumed that ##\hat{A}## is not explicitly time dependent).

So e.g., the HUP for position and energy reads
$$\Delta x \Delta E \geq \frac{\hbar}{2} |\langle \hat{v}_x \rangle|.$$
Here ##\hat{v}_x## is the velocity ##x##-component.

For the usual simple Hamiltonians for one particle
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2+V(\hat{\vec{x}})$$
of course you get
$$\hat{\vec{v}}=\frac{1}{m} \hat{\vec{p}}.$$
Note, however that in general ##\hat{\vec{p}}## is a canonical rather than the kinetic momentum. An important example is the motion of a (spin-0) particle in a magnetic field
$$\hat{H}=\frac{1}{2m} (\hat{\vec{p}}-q \vec{A}(\hat{\vec{x}}))^2,$$
where ##\vec{B}=\vec{\nabla} \times \vec{A}##.
 
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