can one measure energy to a finite accuracy?

Question

Can one measure energy to a finite accuracy in a bounded amount of time? I don't know much about QM, but someone told me that the energy-time uncertainty principle says that it would take an infinite amount of time to ensure that a measurement of energy is inaccurate by no more than a finite value.

As a more concrete instance, can some device in a bounded amount of time find out which energy state an electron of the hydrogen atom is in? Doing so seems to violate Buridan's principle [1], which does not have a quantum mechanical proof or disproof anyway.

[1] : Lamport, L. (2012). Buridan’s Principle. Found Phys 42, 1056–1066. http://link.springer.com/article/10.1007/s10701-012-9647-7

Answer 1

When we think about the state of a hydrogen atom we instinctively think about the solutions to the time independant Schrodinger equation. These are the well known atomic orbitals.

However for the time independant Schrodinger equation to apply the hydrogen atom must have existed unchanged for an infinite time and it will then continue to exist unchanged for an infinite time into the future. The wavefunctions of real hydrogen atoms are technically not the eigenfunctions of the Schrodinger equation. However we could write the wavefunction of a real hydrogen atom as a sum of these eigenfunctions:

$$ \Psi_H = a_{1s}\psi_{1s} + a_{2s}\psi_{2s} + a_{2p}\psi_{2p} + ... $$

where in the real word the coefficients $a_{1s}$ etc would be functions of time. So measuring the energy precisely comes down to measuring the $a$ coefficients precisely.

Suppose we're trying to measure the energy of the ground state. If we wait an infinite time we expect $a_{1s} \rightarrow 1$ and the other coefficients to become zero, and we could measure the energy of the atom to be precisely the energy of the $1s$ eigenfunction. On a finite timescale $a_{1s} < 1$ and the other coefficients would be non-zero so the energy wouldn't be equal to $E_{1s}$.

However, for all but the shortest timescales we're going to find that one of the coefficients is approximately unity and the others are all approximately zero, and that allows us to approximately specify which energy state the atom is in. For example if we found:

$$ \Psi_H = 0.000001\psi_{1s} + 0.999998\psi_{2s} + 0.000001\psi_{2p} $$

Then only the most pedantic of quantum mechanicists would deny that the atom was in the $2s$ state.

It isn't obvious to me how the Buridan's ass argument applies here.

Answer 2

If you know roughly the state of the particle, i.e. you know the energy spectrum of the corresponding wave packet, you can measure that particle (wave packet's energy spectrum) for finite time and then cut off and be sure that you have enough information for finite accuracy in a sense of energy.

But if you do not know the corresponding wave packet, you really need to wait for infinite time that the whole wave packet passes your measurement device, and only then you can be sure that you have measured the whole energy spectrum in desired accuracy.

Actually there is no real time-energy uncertainity principle, but one can set up corresponding idea in various cases. One case is that time-energy measurement accuracy:

We have usual Heisenberg's Uncertainity Principle: $$\Delta x\Delta p \geqq \hbar/2$$ Uncertainity of energy for wave packet in constant potential: $$\Delta E=(\Delta p \centerdot p)/m $$ Now $\Delta t$ is the time that wave packet needs to take for passing measurement device for measuring its energy spectrum: $$\Delta t =\Delta x/v=m\Delta x /p$$ Adding everything up leads to time-energy uncertainity principle: $$\Delta E \Delta t=(\Delta p \centerdot p)/m \centerdot m\Delta x /p =\Delta x\Delta p \geqq \hbar$$