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I'm watching MIT online lectures Quantum Physics I (roughly from one hour mark in the video). The lecturer explains wave functions that describe "Stationary States" that consist of a single energy eigenfunction then he points out that no such thing exists in real life. At some point a student asks a question: "Wait a minute, you said that if we measured the energy of the system, the wave function would collapse into one of the eigenfunctions. So wouldn't measuring the energy of the system would cause it to collapse into a single eigenfunction, thus creating a Stationary State that you just said didn't exist?" The lecturer answers that you can't measure energy with arbitrary accuracy and leaves it at that.

My question is this, though: so what does happen if you measure the energy with some inaccuracy? The wave function does collapse, right? But what does it collapse into? Sounds like it depends on how accurate your measurement was: if you did a crappy job, then it collapses a 'little' (some eigenfunctions are eliminated from the superposition), if you did a good job, it collapses 'a lot'. Am I on the right track or is this completely wrong reasoning? (Either case, it's weird: it's as if the system knew about the accuracy with which a measurement was performed.)

Qmechanic
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Lajos Nagy
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2 Answers2

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Did you cover the uncertainty principle? In quantum mechanics there is and uncertainty between energy and time:

$$ \Delta E \Delta t > \frac{h}{4\pi}$$

this means that if you try to measure Energy with perfect accuracy you will have a great uncertainly in time (actually an infinity uncertainty). I guess this is what the professor was referring to, and he probably did't expand on his answer because you will cover this later on.

now consider an energy eigenstate of the time independent schrödinger equation $\psi(x)$ with eigenvalue E, if you solve the time dependent schrödinger equation for $\psi(x)$ you get the solution:

$$ \psi(x,t) = e^{-iEt2\pi/h}\psi(x) $$

We say the eigenstates evolve in time. So even if you measure with perfect accuracy (which is never possible in quantum meachanics) your state would still not me stationary.

On your question: You are somewhat correct. A wave function will always collapse into a superposition of eigenstates, and if you have some eigenvalues $E_1, \dots, E_n$ of your system and your measurement is very close to some Energy $E_i$, the wavefuction will be still be a superposition but all the coefficients will almost vanishing except for the coefficient of the eigenstate with energy $E_i$.

And yes some aspects of quantum mechanics are strange, but it is the way nature works (we think)

john
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    You should note that energy-time uncertainty is not type of uncertainty principle of position and momentum. Note that the Hamiltonian commutes with $t$, which is just a variable and not an operator. – Gonenc Jul 13 '15 at 09:39
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    For the proper interpretation of the energy-time uncertainty, see this question. – ACuriousMind Jul 13 '15 at 11:42
  • @Paul You probably mean that you start with the time independent Schrödinger equation, where you don't consider the time evolution. But as you move on to the time dependent Schrödinger equation you don't think about states being stationary, you will always try to write you state as a linear combination of eigenstates, for which you know how the time evolution looks like. Or do you mean outside of quantum mechanics? – john Jul 19 '15 at 15:19
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If your measurement would give you an exact upper and/or lower bound, but no more information (i.e. a probability distribution), the state would collapse into the projection onto the subspace of possible values, so it would still be a superposition.

More generally and realistically, we would measure a value, and assign decreasing (classical) probabilities for states with eigenvalue further away from the observed value. This gives us a mixed state: a classical mixture of (pure) quantum states (so we have eigenstates, associated to some observable, pure states, which are quantum superpositions of eigenstates, and these mixed states).

Such states can be conveniently descibed by a density operator.

doetoe
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