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I am watching Susskind's Stanford Lectures on quantum mechanics.

The eigenvectors (eigenfunctions) of the position operator are of the form $\delta(x-k)$. But $$\int\delta^{*}(x-k)\delta(x-k)\, \mathrm dx$$ is undefined or infinite. So as far as I can tell it cannot be the state of a system, since $$\int\psi^{*}(x)\psi(x)\,\mathrm dx = 1$$ is required.

I thought that when we measure position, we could say that the wave function is in the state corresponding to the eigenvector of the eigenvalue measured. But this can't be right because of the above. What can we say about the state of a quantum system right after measuring position?

Is this somehow related to measurements of position not being strictly point-like?

Qmechanic
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Ameet Sharma
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1 Answers1

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Actually, the outcome of the experimental apparatus is an interval $(x_0-\delta, x_0+\delta)$. $\delta>0$ stays for the accuracy of the instrument which can be made smaller and smaller but cannot be removed.

It is therefore assumed (Luders-von Neumann's axiom) that, if the state immediately before the measurement was determined by the wavefunction $$\psi\:,$$ the one immediately after is, up to the normalisation, $$\chi_{(x-\delta, x+\delta)}\psi\:.$$ Here $\chi_{(x_0-\delta, x_0+\delta)}(x)=1$ if $x\in (x_0-\delta, x_0+\delta)$ and $\chi_{(x_0-\delta, x_0+\delta)}(x)=0$ if $x\not \in (x_0-\delta, x_0+\delta)$.

Abstractly, if the initial vector state state is $\psi$, the final one is $P_{(x_0-\delta, x_0+\delta)}\psi$. The operator $P_{(x_0-\delta, x_0+\delta)}$ being the orthogonal projector of the PVM of the position operator associated to the interval $(x_0-\delta, x_0+\delta)$.

Obviously this is the theoretical description of a very ideal measurement procedure.

If, differently from the position observable whose spectrum is continuous, the measurement process regards an observable $A$ with point spectrum and the elements of the spectrum are isolated points, it is always possible to assume the existence of a measurement instrument whose sensibility $\delta>0$ is smaller that the distance of pairs of consecutive values of $A$. This way, even if the measurement is affected by an experimental error represented by $\delta>0$, we can distinguish between couples of eigenvalues and the Luders-von Neumann's axiom takes the more familiar standard form: The state after the measurement with outcome $a_0$ procedure is represented by the eigenvector with eigenvalue $a_0$. (This is true if the eigenspace has dimension $1$ otherwise the general abstract form of L-vN's axiom holds again.)

ADDENDUM (After some discussions with valerio92).

I prove here in details how the general form of L-vN's postulate leads to the standard postulate of collapse of the wavefunction after an imprecise measurement of position.

For a quantum system described over the separable complex Hilbert space $\cal H$, a quantum state is a probability measure $\mu$ over the non-Boolean lattice $\cal L(\cal H)$ of orthogonal projectors in the Hilbert space (see this answer of mine for further details). $P \in \cal L(\cal H)$ has the meaning of an observable whose values are only $0$ or $1$, I mean a YES/NO observable. $\mu(P)$ is the probability that $P$ turns out to be true if measuring it when the state is $\mu$.

Gleason's theorem proves that, if the Hilbert space is separable with dimension $\neq 2$, there is a one-to-one correspondence between these probability measures and density matrices. These are unit-trace, trace class, positive operators $\rho : \cal H \to \cal H$. This correspondence is such that $$\mu_\rho (P) = tr(\rho P)$$ for every $P \in \cal L (\cal H)$.

The general form of L-vN's axiom is the following.

L-vN's axiom. Let $P$ be an orthogonal projector representing an elementary observable of the physical system and $\rho$ a state. If the outcome of the measurement of $P$ when the state is $\rho$ is $1$ (YES), the post measurement state is $$\rho_P = \frac{P\rho P}{tr(\rho P)}\:.\tag{1}$$ This postulate has a direct conditional-probabilty interpretation: $\mu_{\rho_P}$ is the unique probability measure such that $$\mu_{\rho_P}(Q) = \frac{\mu(Q)}{\mu(P)}$$ for every $Q \in \cal L (\cal H)$ with $Q \leq P$.

Pure states by definition are extremal elements of the convex body of the mentioned measures. In other words a density matrix $\rho$ is a pure state if there are no $p,q\in (0,1)$ with $p+q=1$ and density matrices $\rho_1\neq \rho_2$ such that $\rho = p\rho_1 + q \rho_2$.

It turns out that $\rho$ is pure if and only if it has the form $$\rho = |\psi \rangle \langle \psi |$$ for some $\psi \in \cal H$ with $||\psi||=1$.

Evidently unit vectors $\psi$ and $\psi'$ define the same pure state if and only if $\psi = e^{ia}\psi'$ for some $a \in \mathbb R$.

L-vN's postulate applied to pure states specializes to this statement

L-vN's axiom (pure states). Let $P$ be an orthogonal projector representing an elementary observable of the physical system and $|\psi \rangle \langle \psi |$ a pure state. If the outcome of the measurement of $P$ when the state is $|\psi \rangle \langle \psi |$ is $1$ (YES), the post measurement state is still pure has has the form $$|\psi_P \rangle \langle \psi_P | = \frac{P|\psi \rangle \langle \psi | P}{tr(|\psi \rangle \langle \psi | P)}\:.\tag{2}$$

Since $tr(|\psi \rangle \langle \psi | P)= ||P \psi||^2$, so that $$|\psi_P \rangle \langle \psi_P | = \frac{P|\psi \rangle \langle \psi | P}{||\psi|| \: ||\psi||}\:,$$ the postulate can be rephrased as follows.

L-vN's axiom (pure states 2). Let $P$ be an orthogonal projector representing an elementary observable of the physical system and let the unit vector $\psi \in \cal H$ represent a pure state up to phases. If the outcome of the measurement of $P$ when the state is represented by $\psi$ is $1$ (YES), the post measurement state is still pure and is represented, up to phases, by the unit vector $$\psi_P = \frac{P\psi}{||P\psi||}\:.\tag{3}$$

Let us eventually come to the position observable for a particle moving along the real axis.

Here ${\cal H} = L^2(\mathbb R, dx)$ and the position observable is the self-adjoint multiplicative operator $$(X \psi)(x):= x\psi(x)$$ with obvious domain.

The spectral decomposition of $X$ associates it to the projector-valued measure $\{P_E\}_{E \in B(\mathbb R)}$ where $B(\mathbb R)$ is the class of Borel set of the real axis, for instance $E$ could be an interval $E=(a,b)$. The spectral theorem says that $P_E$ is defined like this $$(P_E \psi)(x):= \chi_E(x)\psi(x)\quad \forall x \in \mathbb R\:.\tag{4}$$ The meaning of the orthogonal projector $P_E$ is

$$\mbox{"the position of the particle stays in $E$"}$$

In fact, if $\psi \in L^2(\mathbb R, dx )$ is a normalized vector representing a pure state, the probability that $P_E$ is true is $$tr(P_E |\psi\rangle \langle \psi|)= ||P_E\psi||^2 = \int_{\mathbb R} |\chi_E(x) \psi(x)|^2 dx = \int_E |\psi(x)|^2 dx\:,$$ in perfect agreement with more elementary versions of Quantum Mechanics.

With this interpretation, measuring $X$ means measuring every $P_E$ or at least measuring the class of mutually exclusive elementary propositions $P_{(ns, (n+1)s]}$ with $n \in \mathbb Z$ where $s$ is the sensibility of the instrument.

Let us suppose that the initial state of the particle is pure and (up to phases) represented by the (normalized) wavefunction $\psi$. Suppose that we perform a measurement of the position and we find that the particle stays in $E \subset \mathbb R$. What is the post measurement state according to the Postulate by Luders and von Neumann?

We can apply our third version, the one specialized to pure states described in terms of unit vectors. The post-measurement state $\psi_E$, up to phase, is represented by the vector in (3): $$\psi_E := \frac{P_E \psi}{||P_E \psi||}\:.$$ Taking (4) into account, we find $$\psi_E(x) = \frac{\chi_E(x) \psi(x)}{\sqrt{\int_E |\psi(z)|^2 dz}}\:.$$

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    I'm trying to avoid getting into interpretational questions... but is this $\delta$ purely epistemic? I mean... is it the case that the particle is actually at a certain position but because instruments cannot be infinitely precise, we cannot know exactly? Or is this $\delta$ something more than that? – Ameet Sharma Apr 05 '16 at 10:06
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    It is a very difficult question. In view of the standard interpretation of quantum formalism it should not be epistemic. Also for this reason, the postulate by Luders and von Neumann is suspiciously viewed in the case of continuous spectrum. In this case a more accurate description of the quantum measurement process is perhaps provided by the notion of quantum operation and POVMs and related Kraus operators. – Valter Moretti Apr 05 '16 at 10:13
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    @ValterMoretti Sadly, the picture for observables with continuous spectrum is not nice for repeatability, as far as I know. Let's start with a nice result: for any observable (even with cont spectrum) it is possible to define at least one measurement process $M$, that satisfies the reasonable properties of a measurement. Let's denote by $\mathrm{Ex}_M(I_2\lvert I_1;\rho)$ the conditional expectation of measuring the observable in an interval of values $I_2$, after a first measurement gave a value in $I_1$ (for an initial state $\rho$). – yuggib Apr 05 '16 at 13:54
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    Then $\mathrm{Ex}M(I_2\lvert I_1;\rho)=\mathrm{Tr}[\rho \chi{I_1\cap I_2}]/\mathrm{Tr}[\rho\chi_{I_1}]$ if and only if the observable has purely discrete spectrum (at least within the above - omitted - definition of measurement, that is pretty reasonable). That means essentially that within the usually accepted (von Neumann) measurement scheme, only discrete observables satisfy the L-vN's axiom. The reference for the above results is a paper by Ozawa (1984) – yuggib Apr 05 '16 at 13:58
  • When looking for "Neumann-Lüders postulate/axiom" online I find something that looks quite different from the postulate you are referring to (for example here or here). Could you please provide some literature reference to this version of the postulate? – valerio Nov 02 '16 at 11:31
  • I cannot understand your remark: in both references the postulate has the form I pointed out extended to mixed states. If $\rho$ is the state before the measurement and the outcome of the measurement of an elementary proposition represented by the orthogonal projector $P$ is $1$, then the post measurement state is $P\rho P/ tr(\rho P)$. For pure states $\rho = |\psi x \psi |$ this is equivalent to a post measurement state $P |\psi\rangle/||P |\psi\rangle||$. – Valter Moretti Nov 02 '16 at 11:54
  • If finally $P$ is the proposition associated to the statement "the psition of the particle belongs to the interval $I$", from the spectral decomposition of the position operator, you immediately see that the post measurement wavefunction is, up to its normalization, $\chi_I \psi$ when $\psi$ is the pre-measurement wavefuntion. – Valter Moretti Nov 02 '16 at 11:56
  • See also my answer here http://physics.stackexchange.com/questions/116595/why-is-the-application-of-probability-in-qm-fundamentally-different-from-applica/116609#116609, especially the final part and the addendum 4. – Valter Moretti Nov 02 '16 at 12:13
  • @ValterMoretti You forgot to tag me so I read your comments only now by chance. What I mean is that I cannot find anything about the wavefunction after the measurement being $\chi_I \psi$ in the cited reference, and online in general. Nevermind the mixed state, that is only a generalization: I just don't see why the wavefunction should be in that form, and I would like to have some kind of reference to understand better. – valerio Nov 02 '16 at 15:29
  • Sorry, it is evident from the case of a mixed state, viewing writing it as a density matrix – Valter Moretti Nov 02 '16 at 15:43
  • @ValterMoretti You forgot again to tag me. Anyway, why should it be evident? I stress that I'm not pointlessly arguing but just trying to understand. – valerio Nov 02 '16 at 15:45
  • Sorry, I am being too busy, I will add a comment to my answer – Valter Moretti Nov 02 '16 at 16:06
  • @valerio92 I have added a long addendum to my answer... – Valter Moretti Nov 02 '16 at 17:13
  • @valerio92 Please, let me know if the issue is more clear now or, if it is not, where are the obscure points. I apologize, it was not my intention to be offensive in any sense, I understand how these things may turn out to be a bit difficult the first times one faces them. – Valter Moretti Nov 02 '16 at 17:23
  • Thank you very much, this is terrific! Even if I have to admit that I am not used to such a mathematically rigorous formulation of QM, so I don't fully understand your explanation. But from what I understand, in the case of an observable $A$ with continuous spectrum we need to define the projectors in a "special" way because it is not possible to write the observable as $A=\sum_i a_i P_i$? – valerio Nov 03 '16 at 01:19
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    Yes, if the spectrum is continuous one needs the full version of the spectral theorem which generalize $A = \sum_i a_i P_i$. In this case, projectors are not associated to points $a_i$, but to full intervals $E$ (and also more complicated sets) and the projectors are labeled by these sets $P_E$. The projectors associated to the position operator $X$ are simply given by the characteristic functions $\chi_E$ acting as multiplicative operators. – Valter Moretti Nov 03 '16 at 16:27
  • Great. I think that that was the part I was missing! By the way, my interest came from this question of mine. But now that we discussed this, probably I understand what the answer could be, although I don't know if I am able to properly express it in a formal way. PS Thank you very very much for the addendum: I had already upvoted the answer before you added it, and now I wish I could do more because the addendum is just great! – valerio Nov 03 '16 at 16:59