In non-relativistic QM, the ΔE in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of n identically prepared systems as n goes to infinity. What does the Δt mean, since t is not even an observable?
Let a quantum system with Hamiltonian H be given. Suppose the system occupies a pure state |ψ(t)⟩ determined by the Hamiltonian evolution. For any observable Ω we use the shorthand ⟨Ω⟩=⟨ψ(t)|Ω|ψ(t)⟩.
One can show that (see eq. 3.72 in Griffiths QM) σHσΩ≥ℏ2|d⟨Ω⟩dt|
where σH and σΩ are standard deviations σ2H=⟨H2⟩−⟨H⟩2,σ2Ω=⟨Ω2⟩−⟨Ω⟩2
and angled brackets mean expectation in |ψ(t)⟩. It follows that if we define ΔE=σH,Δt=σΩ|d⟨Ω⟩/dt|
then we obtain the desired uncertainty relation ΔEΔt≥ℏ2
It remains to interpret the quantity Δt. It tells you the approximate amount of time it takes for the expectation value of an observable to change by a standard deviation provided the system is in a pure state. To see this, note that if Δt is small, then in a time Δt we have |Δ⟨Ω⟩|=|∫t+Δttd⟨Ω⟩dtdt|≈|d⟨Ω⟩dtΔt|=|d⟨Ω⟩dt|Δt=σΩ
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