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The way I've understood it is that eigenfunction of an operator are the different states which the actual wavefunction can take when the property/observable corresponding to the given operator is being observed.

For the Hamiltonian operator, the eigenfunctions are forms the wavefunction will take when energy is being measured. The probability that we get a specific state corresponds to the coefficient from fourier decomposition of the wavefunction.

I've also read/interpreted/misinterpreted that upon measurement of position, a wavefunction will collapse to a single spike with a specific value of position (determinate state of the position observable).

With all of the above in mind, is it correct to think that the eigenstates of the position operator are continuous, because the probability density function is continuous and we can get a spike anywhere?

Also, does this mean that the wavefunction takes the form of a spike when position is being observed, but a sine wave when the energy is being observed?

Edit: by continuous eigenstates I want to say that there is an infinite number of them.

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  • $\begingroup$ "With all of the above in mind, is it correct to think that the eigenstates of the position operator are continuous, because the probability density function is continuous and we can get a spike anywhere?" It's not clear why you think the probability density being continuous implies the eigenfunctions are continuous. To figure out if the eigenfunctions of the position operator are continuous, just figure out what the eigenfunctions of the position operator ARE. Then stare at those functions and decide if you think they're continuous! This is independent of the probability distribution. $\endgroup$ – Jahan Claes Nov 11 '18 at 22:28
  • $\begingroup$ I thought that they must be continuous since, from what I understood, eigenstates are what the wavefunction collapses to upon measurement of a given observable. In the case of position - the wavefunction may collapse to any place (within the probability distribution), hence it seems like there are infinite states to which the wavefunction can collapse. Please point out the errors in my reasoning, I feel like I can't advance due to the numerous problems which I can't seem to iron out. $\endgroup$ – Marek Zakrzewski Nov 11 '18 at 22:33
  • $\begingroup$ I agree there are infinite states to which the wavefunction can collapse! I don't see why that implies that any given ONE of those states is continuous? $\endgroup$ – Jahan Claes Nov 11 '18 at 23:35
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    $\begingroup$ @JahanClaes I believe that Marek means to ask "Are the eigenstates of the position operator indexed by a continuous variable?" $\endgroup$ – J. Murray Nov 11 '18 at 23:40
  • $\begingroup$ @J.Murray The edit suggests you are correct! FWIW, Marek, you might want to say something like continuous eigenspectrum to avoid confusion. Eigenstates are functions, which are either discontinuous or continuous as functions. You're not really interested in the eigenstates, you're asking about the eigenvalues or eigenspectrum. $\endgroup$ – Jahan Claes Nov 12 '18 at 0:08
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This is a good question, and the answer depends on how mathematically watertight we want to be.

In the standard physicist's presentation of quantum mechanics, we start with a Hilbert space $\mathcal H$ (e.g. the set of square-integrable functions on a line, $L^2(\mathbb R)$). An observable corresponds to a hermitian (really, self-adjoint) operator $A$ on $\mathcal H$, which has the property that $$\langle \psi, A \phi\rangle = \langle A \psi, \phi\rangle$$

Every hermitian operator $A$ has a set of orthonormal eigenstates $\psi_i$ and corresponding eigenvalues $\lambda_i$, such that $A \psi_i = \lambda_i \psi_i$ and $\langle\psi_i,\psi_j\rangle = \delta_{ij}$

Additionally, this set of eigenstates actually spans the space, so given any arbitrary state $\Psi$, we can expand it like this: $$\Psi = \sum_{i=1}^\infty c_i \psi_i$$

The probability of measuring the observable $A$ to have value $\lambda_i$ is now equal to $|c_i|^2$, and if we perform the measurement and get $\lambda_i$ as a result, the state "collapses" into the corresponding $\psi_i$.

When we apply this to the position operator $X$ (which acts by multiplying wavefunctions by the variable $x$), we say that the eigenstates of $X$ are the delta functions centered at all real numbers, because

$$ x \delta(x-x_0) = x_0 \delta(x-x_0)$$

Therefore, if we measure the position of a particle and get a real number $x_0$, the state collapses into the state $\delta(x-x_0)$, which is an infinitely sharp spike centered at $x=x_0$.


Everything I said (other than sweeping the distinction between self-adjoint and hermitian under the rug) is actually mostly accurate up until I mentioned the position operator. The problem is that the formalism described up to that point only applies to operators with discrete spectra.

The position operator has a purely continuous spectrum - namely $\mathbb R$, for a particle on a line. The problem with this is that such operators actually don't have any eigenvalues or eigenstates. Delta functions don't count because they aren't actually states in $\mathcal H$ - they are not square-normalizable functions (or indeed, functions at all).

The notion of a spectrum is more general than that of eigenvalues. For an operator which only has a discrete spectrum, the spectrum is simply equal to its set of eigenvalues. However there are many operators (position, for example) which have purely continuous spectra, and therefore no eigenvalues at all. There are even operators which have a combination of both, like the energy operator in the case of a hydrogen atom. The spectrum of that operator is discrete for negative energies (corresponding to bound states) but continuous for positive energies (corresponding to scattering states).

To understand operators which have continuous spectra, we need the machinery of the spectral theorem, which would take the better part of a lecture course on functional analysis to develop properly. Instead, I'll just directly address your question.


Let $\psi$ be a wave function. The spectrum of $X$ is just the real line, so consider any interval $I\subseteq \mathbb R$. The spectral projector $P_I$ maps an unmeasured state to the state you'd have after you measured the position to be in the interval $I$.

In the case of the position operator, the appropriate projector is just the indicator function: $$ P_I = \mathbb 1_I(x) = \cases{1 & $x\in I$\\0 & $x\notin I$}$$

I'll be concrete. Let's say that the wavefunction of my particular state is a Gaussian, so $\psi(x) = e^{-x^2}$ enter image description here

Now I devise an experiment to measure whether the particle is in the interval $[\frac{1}{2},1]$. If I perform my measurement and find that my particle is in that interval, then the resulting projector is the function

$$ P = \cases{1 & $x\in[\frac{1}{2},1]$\\0 & elsewhere}$$ and my new state would be this:

enter image description here

On the other hand, if I perform my measurement and find that my particle is not in that interval, then the resulting post-measurement state would be this:

enter image description here

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  • $\begingroup$ Thank you so much for your answer! Can I also ask you if it'd be correct to say that the wavefunction of a particle will turn into a sine wave upon measurement of its energy? $\endgroup$ – Marek Zakrzewski Nov 12 '18 at 0:07
  • $\begingroup$ That depends entirely on the system. For an infinite potential well, the energy operator has a purely discrete spectrum, and so it has a complete set of eigenstates which happen to be sine functions ($\psi_n \propto \sin\left(\frac{n\pi x}{L}\right)$). In that case, the answer is yes. For the harmonic oscillator, the same is true but the eigenstates are the product of a Gaussian with a hermite polynomial (see here), so measurement of the energy will result in a collapse into one of those. $\endgroup$ – J. Murray Nov 12 '18 at 0:11
  • $\begingroup$ For the free particle on a line, the energy operator has a purely continuous spectrum and, like the position operator here, has no eigenstates in $L^2(\mathbb R)$. This case also requires the machinery of the spectral theorem to fully understand. $\endgroup$ – J. Murray Nov 12 '18 at 0:12

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