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I'm reading a book on quantum mechanics and a topic in there was not well explained. I couldn't find an answer on the web.

Say $\psi(x)$ is a position wave function of a one dimensional free particle, and $\hat{\psi}(k)$ the momentum wave.

Say $\psi(x)$ and $\hat{\psi}(k)$ are the position and momentum space representation of wave function respectively, for a free particle moving in one dimension.

We'd like to perform a measurement to check if the particle's position is inside some range $I=(a,b)$. So they do the following: Let $\chi_I(x)$ be a characteristic function such that $\chi_I(x)=1$ if $x\in I$ and $\chi_I(x)=0$ otherwise. And we'll define a linear operator $P_I:\psi \to \chi_I \psi$.

The book says $P_I$ is a projection operator that corresponds to an ideal measurement device that makes a measurement that can result in either a "yes" or a "no". The probability for "yes" is $\langle \psi(x),\chi_I \psi(x) \rangle$ If the result is "yes", the position wave collapses to $\chi_I \psi$, otherwise it collapses to $(1-\chi_I) \psi$.

Is this correct? I didn't see this kind of formalism for continuous properties anywhere else.

The book also says an operator like this can be defined for the momentum wave function. I'm guessing this would be applying $\chi_I$ to $\hat{\psi}(k)$. But how would such an operator be described as a linear operator on $\psi(x)$? I.e., what would be a linear operator on $\psi(x)$ that corresponds to a measurement device that tells us if the momentum is within range $I=(a,b)$?

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Notice that the probability of measuring say the position of a particle whose wavefuction is $\psi(x)$ in the interval $I=(a,b)$ is $$\int_a^b \left|\psi(x)\right|^2 \mathrm{d}x.$$

We can define a multiplication operator on the state space much like the position operator $\hat{X}\psi(x)=x\psi(x)$ as follows.

$P_I(\psi)=\chi_I(x)\psi(x)$. It is a projection since $\chi_I(x)^2=\chi_I(x)$ for all $x$, since $0^2=0$ and $1^2=1$. So $P_I^2(\psi)=P_I \psi$, then taking the $L^2$ inner product gives:

$$\langle\psi,P_I \psi\rangle=\int_{-\infty}^\infty \psi(x)^*\chi_I(x)\psi(x)\mathrm{d}x=\int_a^b \left|\psi(x)\right|^2 \mathrm{d}x$$

So it is in fact the measurement as mentioned above.

The measurement that is being performed here is "is the particle somewhere between $a$ and $b$", of which the outcomes are "yes" or "no". If yes then by the postulates of measurement the wave function collapses to $$\frac{P_I \psi}{\sqrt{\strut\langle\psi,P_I\psi\rangle}}=\frac{\chi_I(x)\psi}{(\int_a^b \left|\psi(x)\right|^2 \mathrm{d}x)^{1/2}}$$

so that the result is properly normalised.

If the result was "no" then the state would project onto the complementary subspace which would be given by $1-P_I$ which is also a projection operator. Thus the state collapses to:

$$\frac{(1-P_I) \psi}{\sqrt{\strut\langle\psi,(1-P_I)\psi\rangle}}=\frac{(1-\chi_I(x))\psi}{(1-\int_{a}^b \left|\psi(x)\right|^2 \mathrm{d}x)^{1/2}}$$

again properly normalised, and using $\langle\psi, \psi \rangle=1$.

Yes you could also define $\hat{P}_I$ for $\hat{\psi}(k)$ in a similar way where $I$ would now be an interval in momentum space.

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  • $\begingroup$ Thanks a lot, that helped. But regarding the operator for momentum, can $\hat{P}_I$ be defined instead as a linear operator on $\psi(x)$? For example, would $FT^{-1} \hat{P}_I FT$ (Where FT is the Fourier transform) be a linear operator on $\psi(x)$ that measures whether the momentum is in range I? $\endgroup$ – Udi256 Jul 13 '16 at 12:59
  • $\begingroup$ @Udi256 I think you might be able to write the operator as a convolution in position space - en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem . As in $\hat{P}_I \hat{\psi}$ in position space would look like $\int_{-\infty}^\infty \mathrm{FT}^{-1}(\chi_I)(x) \psi(x-y)\mathrm{d}y$ probably up to some factors of $2\pi$ depending on fourier transform conventions. $\endgroup$ – snulty Jul 13 '16 at 13:24
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$\chi_I$ is a number, not an operator, do not mix up $\chi_I$ with the projection operator. This whole formalism is actually very simple in this case.

$P_I$ is a projection operator onto the space of states which correspond to a particle being inside the interval $I$. Suppose we live in discrete space for the time being so we can work with explicit representations of $\psi$.

Let's say we live on a lattice with 10 sites labeled 1-10, and let $I = [2, 5]$. The wave function for a particle in position space is

$\psi = a_1 |1> + a_2 |2> + ... + a_{10} |10>$,

where the $a_i$ are the amplitudes to find the particle at position $i$.

As an aside, there is no difference between $\psi$ and $|\psi>$ here, it's just convenient looking to wrap $\psi$ in brackets to denote expectation values. You can equally think about a 10 component column vector where each component is an $a_i$. These are all the same objects.

Given the notation used above, the projection operator takes any $\psi$ and shoves 0 into all the components that don't correspond to $\{|2>,|3>,|4>,|5>\}$ while leaving the other components the same. An explicit way to write this is

$P_I = 0*|1><1| + 1*|2><2| + 1*|3><3| + ... + 0 |10><10|$.

You could work out the matrix form as well in the case of using a column vector to represent $\psi$.

Generally in quantum mechanics, measurements can be described by such operators. The action of a measurement on a wavefunction is

$ |\psi> \rightarrow \hat{O} |\psi> $

where $\hat{O}$ is a projection operator corresponding to some measurement. This explicitly shows the "collapse" of the wave function, since after the measurement, we know that states which give other measurement outcomes could not be the state of the particle. This is reflected by the fact that $\psi$ has components not compatible with the measurement (in this case, finding the particle in the interval) turned to 0 after being acted on with $\hat{O}$.

The Born Rule states that the expected value of a measurement on a pure state $\psi$ is

$<\psi| \hat{O} |\psi>$

for a measurement operator $\hat{O}$, this is what you were trying to quote when you gave the probability of "yes". But what you wrote was wrong because you did not act on $\psi$ with a measurement operator instead you just multiplied $\psi$ by a number.

The piece of the wavefunction which corresponds to a yes measurement is the projection of the state onto the points in space in the interval.

The continuous case is completely analogous, except now we can not write a nice column vector since it would have infinite components. Instead of having a discrete set of $a_i$, we instead take a function $\psi(x)$ over a continuous set of values $x$.

However, the fundamental ideas are more clear in the discrete case, and all the algebraic properties can be seen much more clearly that way.

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    $\begingroup$ Pro tip: you can make bras and kets look like $|1\rangle$ and $\langle 2|$ using $|1\rangle$ and $\langle 2|$. $\endgroup$ – Emilio Pisanty Jul 13 '16 at 12:53
  • $\begingroup$ The only possible outcome of $P_I$ here is that the particle in inside $I$. Can you write an operator that represents the whole experiment by also including the outcome where the particle is outside $I$? Or do you always have to use either $P_I$ or $Id-P_I$ depending on the outcome ($Id$=identity matrix) $\endgroup$ – user3502079 Dec 20 '17 at 10:10
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For any self-adjoint operator $A$ (with continuous or discrete spectrum), the so-called spectral theorem tells you there is unique a family $(P_{\lambda}(A))_{\lambda\in\mathbb{R}}$ of orthogonal projections (called spectral family) such that $P_{\lambda_2}(A)-P_{\lambda_1}(A)$ gives you the projection on the subspace with spectrum in the interval $[\lambda_1,\lambda_2]$ (of course I am assuming here $\lambda_2>\lambda_1$).

Another common notation is $\chi_{[\lambda_1,\lambda_2]}(A)=P_{\lambda_2}(A)-P_{\lambda_1}(A)$. Of course this coincides with the usual characteristic function for the position.

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protected by Qmechanic Jul 13 '16 at 13:10

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