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As I understand it, there are two "versions" of the Heisenberg uncertainty principle:

Position-Momentum uncertainty \begin{equation} \sigma_x \sigma_p \geq \frac{\hbar}{2} \end{equation}

where $[\hat{x},\hat{p}] = i\hbar$ implies no quantum state can be both a position and momentum eigenstate.

and then

Time-Energy uncertainty \begin{equation} \sigma_T \sigma_E \geq \frac{\hbar}{2} \end{equation}

I don't understand why time and space are separated. Like why isn't $\hat{x}$ an operator that represents information about position in spacetime. We could call this operator $\hat{s}$. Presumably if there is a $\sigma_T$, there must be a time operator $\hat{T}$. If so, why is there a time operator devoid of any reference to spacetime?

I read this 2010 article by John Baez suggesting not everyone agrees that the time-energy uncertainty version is valid. But isn't that one of the ideas that let's us believe virtual particles can exist?

My Question

There were several rhetorical questions here in the sense I intended them as food for thought. The one I want answered is Why isn't the Heisenberg uncertainty principle stated in terms of spacetime?

Stan Shunpike
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2 Answers2

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The problem of including time as an operator rather than a parameter in Quantum Mechanics is what led to the development of Quantum Field Theory. I.e., the position operator was demoted to a parameter rather than promoting time to an operator.

The two uncertainty principles you quote are entirely different. The first (position/momentum) principle is the Heisenberg uncertainty principle. It is fundamental in Quantum Mechanics and follows from the commutation relations you quoted. The second (energy/time) "principle" basically just comes from approximations in rate/scattering theory. It is not "as fundamental".

hft
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I think the question about "time" as an operator is quite controversial. It is not generally accepted that time can be associated to a (possibly unbounded) self-adjoint operator. One could naively argue that the time occurs in physical laws as a parameter rather than something we actually observer, although there clearly are instruments that can measure "time". Recall now that Heisenberg's uncertainty principle comes from postulating non-commutativity of incompatible observables, hence $[A,B] = iC$ leads to $$\Delta_\omega(A)\Delta_\omega(B) \geq |\omega(C)|$$ for any state $\omega$. Since there is no time operator, it is not clear where the energy-time uncertainty relation is coming from, so it is not a Heisenberg-like uncertainty relation, given these premises.

Phoenix87
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  • How did it first come about? I presume someone saw $p\rightarrow E$ and $-i\partial_x\rightarrow i\partial_t$ and suggested an uncertainty principle? – Ryan Unger Feb 03 '15 at 23:45
  • Could be, but apart from this weak argument I don't see how else to justify it without a pair of operators and a commutation relation between them. – Phoenix87 Feb 03 '15 at 23:49
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    what instrument observes "time"? – hyportnex Feb 03 '15 at 23:52
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    @user31748 a "clock" – Robin Ekman Feb 04 '15 at 01:00
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    @Robin Ekman yeah, I know, but surely it ain't the same way as a voltmeter measures voltage or a galvanometer measures current. – hyportnex Feb 04 '15 at 01:51
  • Geiger counters and photographic plates rely on a negative temperature unstable equilibrium in order to amplify their interaction with the microscopic particle up to the level where we can see it. This is essential to the "instrument" that does the measurement. A clock does not interact with a microscopic system in the same way.....that is @hyportnex 's point, as I see it. IMHO, it is decisive. – joseph f. johnson Nov 26 '15 at 21:35