I am self-studying from a text in QM. Well defined states are mentioned several times. By and large these are consistent and seem to be readily apparent:

  • states of well defined energy are basis kets.

  • if $Q$ is an observable, any arbitrary ket can be represented as a linear combination of states {$q_i$} in which values of $Q$ are well defined.

  • whenever $[Q,H] = 0$, a state of well defined $Q$ evolves into another such state.

So far so good. Then comes the "bombshell."

The text later on says:

States of w-d energy are unphysical and never occur in Nature. They are incapable of changing in any way, and hence it is impossible to get a system into such a state.

I would appreciate any explanations reconciling these two apparently contradictory characterizations of well defined states. And, in particular, how then to understand, e.g., a creation operator in the case of a harmonic oscillator in a stationary state.

EDIT Hopefully there might be any answer regarding the last part of the question regarding harmonic oscillators. From the perspective of a naive beginner, it looks like the analysis is a representation of reality.

As I think about this, one thing that comes to mind is that the creation and annihilation operators are not Hermetian, with my presumed inference that this undermines the "reality" of the model.

But in view of the highlighted quote, it seems that it is but an exquisite mathematical exercise. To the extent this is accurate, then what is the relevance of studying it. I would hazard a guess that it might be a framework for studying the anharmonic oscillator.

  • $\begingroup$ To highlights quote: in editing mode, select the block of text then click the "quotes" button. This will highlight. Same is achieved by putting > at the start of every line (clear line before and after) $\endgroup$
    – Floris
    Apr 19, 2014 at 18:33
  • $\begingroup$ The claim "states of well-defined energy never occur in Nature" is true, because "state" is used in the sense "ket". Kets are mathematical concepts introduced to describe spontaneous evolution of microscopic systems. Kets are not some real things that could "occur in Nature". $\endgroup$ Apr 19, 2014 at 20:12
  • $\begingroup$ The claim can be also understood differently: that physical systems cannot have definite energy. I do not think there is much evidence for that. $\endgroup$ Apr 19, 2014 at 20:14
  • $\begingroup$ Which reference? $\endgroup$
    – Qmechanic
    Apr 25, 2014 at 18:45
  • $\begingroup$ @Qmechanic Sorry I must have missed your comment and just saw it now. It's from J. Binney's (Oxford) notes: www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf $\endgroup$
    – user41976
    May 1, 2014 at 16:10

2 Answers 2


The statement you quote is correct and slightly profound till you understand it well enough that it becomes simple :-)

If a system is in an energy eigenstate, then it must exist in that state for all time -- from $t \rightarrow - \infty$ to $t \rightarrow + \infty$.

A few comments:

  1. Clearly, any physical state you create in a lab does NOT have this property. It was created at a finite time and will be destroyed (observed) at a finite time. So any state you might create in a lab cannot be an exact energy eigenstate. It might be one, to a very good approximation. Loosely, a naive "energy-time" uncertainty relation tells you that the longer the state survives, the more definite its energy. (Energy-Time uncertainty in QM is a tricky thing, but think of the analogy with the accuracy in specifying the frequency of a wave compared to the time over which you observe the wave).

  2. When deriving the eigenstates of some system, you assume that the system is isolated. But, to deal with any system in the lab, you have to interact with it in some manner. In such a case, the energy eigenstates of the coupled system are no longer the energy eigenstates of the isolated system.

With both those caveats, energy eigenstates still form a reasonably convenient basis for studying the system. You could in principle take the eigenspectrum of any suitable operator as a basis for your linear vector space, but more often than not, those states evolve in time (since you don't look at the system only for an instant) -- hence, energy eigenstates form a very convenient basis. If you ever have to study states which you actually create/measure in the lab, then you can consider them to be superpositions of energy eigenstates. Since te structure of quantum mechanics is linear, all the analysis you might want to do proceeds in a fairly straight-forward manner.

  • 1
    $\begingroup$ "If a system is in an energy eigenstate, then it must exist in that state for all time -- from $t→−∞$ to $t→+∞$". Could you explain why? Wave function is associated with definite energy if it oscillates as $e^{-i\omega t}$. It suffices that it oscillates in this way for finite time. $\endgroup$ Apr 19, 2014 at 20:19
  • $\begingroup$ Naively speaking, such a wavefunction (evolution) is continuous but not not smooth, so that signals something very weird. More seriously, I think you're imagining an analog with particle in a box, whose Hamiltonian is quadratic $H = \partial_x^2$. So an oscillator over a finite duration continuously connected with "zero" wavefunction is an energy eigenstate with an eigenvalue of $-1$ i.e. a phase shift of $\pi$. On th eother hand, $H \sim \partial_t$ is not quadratic, so it shifts phase of the oscillation only by $\frac{\pi}{2}$ -- and your wavefunction is NOT an eigenstate of this operator. $\endgroup$
    – Siva
    Apr 21, 2014 at 16:51
  • $\begingroup$ Yes, but I do not see any connection to my question. Operator $\partial_t$ is just different operator operating on different parameter. Eigenfunction of some operator $H_0$ operating on $x$ does not need to be eigenfunction of $\partial_t$ operating on $t$. $\endgroup$ Apr 21, 2014 at 21:41
  • $\begingroup$ $H$ is the generator of time translations. So the explanation in my comment above applies to the time profile of a wavefunction $\psi(x,t)$ which is an energy eigenstate of the Hamiltonian: $H \psi(x,t) = E \psi(x,t) \implies \partial_t \psi(x,t) = E \psi(x,t)$. $\endgroup$
    – Siva
    Apr 21, 2014 at 22:46
  • $\begingroup$ Are you sure you did not want to write $\partial_t \psi = \frac{1}{i\hbar}E\psi$? Anyway, I still do not see any ground for the quoted claim. $\endgroup$ Apr 22, 2014 at 5:53

Here are some quotes from A. Peres, "Quantum Theory: Concepts and Methods" which seem quite relevant to me, which I saw in a review on Amazon.

Quantum phenomena do not occur in a Hilbert space, they occur in a laboratory". (Preface)

The essence of quantum theory is to provide a mathematical representaion of >states (that is, of preparation procedures), together with rules for computing >the probabilities of the various outcomes of any test. (p. 26)

The only meaning of "quantum state" is: a list of the statistical properties of >an ensemble of ideally prepared systems. (p. 183)


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