<p>I don't know if this is an "explanation", but hopefully it is a useful "description".</p>

<p>More generally than projective measurements, one always measures an <em>operator</em>. (A projector is a special case of this.) So what does it mean to "measure an operator"?</p>

<p>Well, operators often correspond to 'observable' physical quantities. The most important in quantum mechanics, for instance, is energy; but one can also (sometimes indirectly) measure other quantities, such as angular momentum, <em>z</em>-components of magnetic fields, etc. What is being measured always gives real-valued results --- in principle, some definite result (e.g. an electron is in the  'spin +1/2' state as opposed to 'spin &minus;1/2', or in the first excited energy level as opposed to the ground-state in a hydrogen atom, etc.), albeit each <em>a priori possible</em> result is realized with some probability.</p>

<p>We assign each of the real-valued outcomes of a measurement to a subspace. The way we do this is to describe a Hermitian operator --- <em>i.e.</em> an operator which associates a real eigenvalue to different subspaces, with the subspaces summing up to the whole Hilbert space. A projector is such an operator, where the real values are 0 and 1; <em>i.e.</em> describing that a vector belongs to a designated subspace (yielding a value of 1), or its orthocomplement (yielding a value of 0). These Hermitian operators are <strong>observables</strong>, and the eigenspaces are those for which the observable has a "definite" value.</p>

<p>But what about those vectors which are not eigenvectors, and do not have "definite" values for these observables? Here is the non-explaining part of the description: we project them into one of the eigenspaces, to obtain an eigenvector with a well-defined value. Which projection we apply is determined at random. The probability distribution is given by the familiar Born rule:</p>

<p>$$ \Pr\limits_{|\psi\rangle}\bigl( E = c \bigr) \;=\; \langle \psi | \Pi_c | \psi \rangle \;, $$</p>

<p>where $\Pi_c$ is the projector onto the <em>c</em>-eigenspace of an 'observable quantity' <em>E</em> (represented by a Hermitian operator $A = \sum_c \; c \cdot \Pi_c$). The post-measured state is <em>some</em> projection of the state $|\psi\rangle$ onto <em>some</em> eigenspace of the observable <em>A</em>. And so if $| \psi_0 \rangle$ is the pre-measurement state, $| \psi_1 \rangle$ is the post-measurement state, and $\Pi_c$ is the 'actual result' measured (<em>i.e.</em> the eigenspace onto which the pre-measurement state was actually projected), we have the proportionality result</p>

<p>$$ | \psi_1 \rangle \;\propto\; \Pi_c | \psi_0 \rangle $$</p>

<p>by the projection rule just described. This is why there is the projector in your formula.</p>

<p>In general, the vector $| \psi'_1 \rangle = \Pi_c | \psi_0 \rangle$ is not a unit vector; because we wish to describe the post-measurement state by another unit vector, we must rescale it by</p>

<p>$$ \|\;|\psi'_1\rangle\;\| = \sqrt{\langle \psi'_1 | \psi'_1 \rangle} = \sqrt{\langle \psi_0 | \Pi_c | \psi_0 \rangle} \;,$$</p>

<p>which is the square-root of the probability with which the result would occur <em>a priori</em>. And so, we recover the formula in your question,</p>

<p>$$ | \psi_1 \rangle \;=\; \frac{\Pi_c | \psi_0 \rangle}{ \sqrt{ \langle \psi_0 | \Pi_c | \psi_0 \rangle }} \;.$$</p>

<p>(If this formula seems slightly clumsy, take heart that it looks and feels a little bit better if you represent quantum states by density operators.)</p>

<p><strong>Edited to add:</strong> the above should not be construed as a description of POVMs. A "positive operator valued measurement" is better seen as describing the <em>expectation value</em> of various measurable observables <em>E<sub>c</sub></em> in a collection {&nbsp;<em>E<sub>c</sub></em>&nbsp;}<sub><em>c</em>&nbsp;&isin;&nbsp;C</sub>&nbsp;.</p>

CC BY-SA 2.5
2010-09-09T12:30:20.857Z
2010-09-09T12:57:32.900Z
2010-09-09T12:57:32.900Z
Niel de Beaudrap
<p>If you haven't seen them already I highly recommend Scott Aaronson's lecture notes <a href="http://www.scottaaronson.com/democritus/">"Quantum Computing Since Democritus"</a>, especially <a href="http://www.scottaaronson.com/democritus/lec9.html">lecture 9</a>. They really helped me as a non-expert and I have tried to distill his presentation to the main points <a href="http://rweba.livejournal.com/429528.html">here</a> and <a href="http://rweba.livejournal.com/429744.html">here</a>.</p>

<p>As far as your specific query I think it helps build intuition if you can calculate some simple example's using the Born Rule and see how the Measurement Postulate works in practice.</p>

<p>I found it easiest to think of as "The probability of measuring the ith outcome is the square of the amplitude of the ith element of the state vector - if you do a change of basis to to the eigenvectors of the Operator."</p>

<p>This also ties in neatly with the intuition that quantum mechanics is probability with complex numbers - since the squares of the amplitudes must sum up to 1.</p>

<p>As long as you're studying quantum computing you might also want to check out this discussion of <a href="http://www.scottaaronson.com/blog/?p=208">Shor's algorithm</a>.</p>

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2010-09-11T13:37:05.620Z
Thanks to you Mugizi...Scott Aaronson's lecture notes seem really nice.
Akash Kumar
CC BY-SA 2.5
2010-09-10T09:06:50.293Z
2010-09-10T09:06:50.293Z
Mugizi Rwebangira
<p>Niel de Beaudrap's answer regarding Kraus Operators was very good.  With regard to the Nielsen and Chuang text book, this means that one should read Chapter 2, then Chapter 8, and <em>then</em> the intervening chapters. </p>

<p>Moreover, the Kraus operator representation has an infinitesimal limit called a Lindbladian operator; broadly speaking, Lindbladian operators are to Kraus operators what a Lie algebra is to a Lie group.  Carlton Caves' on-line notes "Completely positive maps, positive maps, and the Lindblad form" cover much of this material.  </p>

<p>The advantage of working exclusively with infinitesimal Lindbladian operators instead of Kraus operators is that the Lindbladians pullback naturally onto non-Hilbert quantum state-spaces; these include the tensor network state-spaces that are becoming ubiquitous in quantum chemistry and condensed matter physics; moreover pullback techniques are ubiquitous in string theory too.</p>

<p>There is at-present no textbook that develops this geometric, non-Hilbert description of quantum dynamics ... but there should be!  Textbooks that (with the above references) in aggregate cover cover the main ideas are John Lee "Smooth Manifolds", Frenkel and Smit "Understanding Molecular Simulation: from Algorithms to Applications", and Kloeden and Platen "Numerical Solution of Stochastic Differential Equations."</p>

<p>It's true that this is a lot of reading ... and this is why geometric quantum dynamics is not taught at the undergraduate level.  This is a pity, because it is all too easy for undergraduates to acquire the fixed notion that the state-space of quantum dynamical systems is a linear vector space, even though this is not true in most large-scale practical calculations.  </p>

<p>As for the state-space that Nature uses: no one knows&mdash;the experimental evidence for local (tangent-space) quantum linearity is fairly strong, yet the evidence for global (Hilbert-space) quantum linearity is fairly weak.  In particular, high-precision molecular beam quantum dynamical experiments&mdash;which many textbooks hold forth as evidence of quantum linearity&mdash;can be simulated with the required relative precision of ~1/2^{65} on low-dimension tensor network state-spaces, with near-perfect dynamical symplecticity replacing near-perfect dynamical linearity.</p>

<p>For the above reasons, perhaps 21st century students should not accept the textbooks of the 20th century entirely at face value.  But really, what 21st century student would want it any other way?</p>

<p>The above is how quantum systems engineers have come to embrace a mathematical toolset that melds geometric and algebraic naturality, and applies generally to classical, quantum, and hybrid dynamical systems.</p>

<hr>

<p><em>Edit addition:</em> As a test of the feasibility of a geometric approach to practical quantum simulation, our Quantum Systems Engineering (QSE) Group supplemented Charlie Slichter's classic textbook <em>Principles of Magnetic Resonance</em> with an enhanced version of Chapter 3 "<a href="http://faculty.washington.edu/sidles/ENC_2011/index.html#Slichter" rel="nofollow noreferrer">Magnetic Dipolar Broadening and Polarization Transport in Rigid Lattices</a>".  </p>

<p>This geometric transcription points naturally to multiple open questions in geometric dynamics; see for example the MathOverflow question "<a href="https://mathoverflow.net/questions/66506/in-quantum-dynamical-simulations-what-is-the-symmetric-riemannian-analog-of-a">In quantum dynamical simulations, what is the symmetric (Riemannian) analog of a Poisson bracket?</a>"</p>

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2010-09-27T11:24:03.553Z
I've seen you wave the flag for this approach all over the net. With a suggestive sentence or two, could you give an idea of how the state spaces you mention are non-linear? With geometric quantization you start with a manifold M as the classical phase space but the quantum state space is the Hilbert space L^2(M). That is, even if the classical geometry is highly non-linear the quantum geometry is still linear, though it is of course much bigger (it has infinite dimension and so on).
Per Vognsen
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2010-09-27T11:30:49.763Z
Sorry, I told a white lie. You actually have to look at L^2 over a line bundle on M. But the basic point remains.
Per Vognsen
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2010-09-27T14:33:03.983Z
Per, what you say is true of the classic (mainly Russian) school of "geometric quantization", in which one starts with a classical system and seeks a quantum generalization of it.   But exactly the <i>opposite</i>  happens in Ashtekar/Schilling models of "geometric quantum mechanics", in which the starting point is symplectic/Lindbladian dynamics on a K&auml;hler manifold.
John Sidles
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2010-09-27T14:50:56.227Z
Hmmm ... let's format it better!   Per, in the (mainly Russian) school of "geometric quantization" one starts with classical dynamics and seeks a quantum generalization of it.   The opposite move is seen in Ashtekar/Schilling models of "geometric quantum mechanics", in which the start is symplectic/Lindbladian dynamics on a Kahler state-space, following which one: (1) exhibits classical dynamics as a limit induced by Lindblad flow, and/or (2) pulls-back onto Hilbert space as a large-N (spectral) approximation.  In engineering, the latter two methods are commonly used, yet not commonly taught.
John Sidles
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2010-09-27T11:07:19.170Z
2011-06-01T17:23:17.137Z
2017-04-13T12:57:54.777Z
John Sidles
<h2>Addendum.</h2>

<p>After re-considering the form of your question (<em>e.g.</em> the M<sup>&dagger;</sup>M in the denominator --- as opposed for instance to a single operator M, which suffices for projectors) and reconsulting my copy of Nielsen and Chaung, here are some supplementary details not covered by my previous answer. (I'm posting this as a separate answer due to length, and because I feel that this is even less of an 'explanation' than my previous answer.)</p>

<p>Suppose that our only means of measuring a qubit <em>X</em> is indirect: by a 'weak' interaction with an ancilla <em>A</em>, followed by a measurement on <em>A</em>.
We would like to be able to talk about these as being in a sense a way of measuring <em>X</em>. How might we describe such a measurement in terms of <em>X</em> alone? Well: suppose we can easily prepare <em>A</em> in the initial state $|+\rangle \propto |0\rangle + |1\rangle$, and perform a controlled unitary of the following sort, with <em>X</em> as the control and <em>A</em> as the target:</p>

<p>$$ U \;=\; \left[\begin{matrix} \quad1\quad &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \quad1\quad &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; \cos(\tfrac{\pi}{12}) &amp; \sin(\tfrac{\pi}{12}) \\ 0 &amp; 0 &amp; -\sin(\tfrac{\pi}{12}) &amp; \cos(\tfrac{\pi}{12}) \end{matrix}\right] $$</p>

<p>We then measure <em>A</em> in the standard basis (so that <em>A</em> now stores the measurement result). This transforms the state of <em>X</em> as follows:</p>

<p>$\begin{align*} |\psi_0\rangle_X \;=&amp;\; \alpha|0\rangle_X + \beta|1\rangle_X
\\\\\mapsto&amp;\; \alpha |0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr) \quad+\quad \beta |1\rangle_X \otimes \bigl(\tfrac1{\sqrt 2}|0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr)
\\\\\mapsto&amp;\; \alpha |0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle_A + \tfrac1{\sqrt 2}|1\rangle_A\bigr) \quad+\quad \beta |1\rangle_X \otimes \bigl(\tfrac{\sqrt 3}2|0\rangle_A + \tfrac12|1\rangle_A\bigr)
\\\\=&amp;\; \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle_X + \tfrac{\sqrt3\beta}{2} |1\rangle_X \bigr) \otimes |0\rangle_A \quad+\quad \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle_X + \tfrac{\beta}{2} |1\rangle_X \bigr) \otimes |1\rangle_A
\\\\\mapsto&amp;\;
\begin{cases}
|\psi_1\rangle_X \otimes |0\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle_X + \tfrac{\sqrt 3\beta}{2}|1\rangle_X\bigr) \otimes |0\rangle_A &amp; \;\text{for the result 0; or } \\
|\psi_1\rangle_X \otimes |1\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle_X + \tfrac{\beta}{2}|1\rangle_X\bigr) \otimes |1\rangle_A &amp; \;\text{for the result 1.}
\end{cases}
\end{align*}$</p>

<p>In the equations above, note that if the result of the measurement is <em>c</em>, the final state $|\psi_1\rangle$ of <em>X</em> is proportional to $|\psi'_1\rangle = M_c |\psi_0\rangle$, where we define</p>

<p>$$ M_0 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{\sqrt 3}{2} |1\rangle\langle 1|\;,\qquad 
M_1 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{1}{2} |1\rangle\langle 1|\;;$$</p>

<p>and we may verify that the probabilities with which we obtain the measurement results are in each case $\langle \psi'_1 | \psi'_1 \rangle \;=\; \langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle$. </p>

<p>This is very close to describing the transformation of <em>X</em> in the same way that we describe projective measurements. But is this any sort of measurement, meaningfully speaking? Well: if we can do statistics on the results of multiple iterations of this procedure, and if <em>X</em> is initially in the standard basis, we would notice that there is a bias in when we obtain the '0' result: we obtain it more often when <em>X</em> is initially in the state $|1\rangle$.
If we can sample enough times to distinguish whether the measurement results are distributed more like $(\frac12,\frac12)$ or $(\frac34,\frac14)$, we can determine with high probability whether the qubit is initially in the state $|0\rangle$ or the state $|1\rangle$.</p>

<p>The similarity of the probabilities-and-update formulae to those of projective measurement, and the fact that we can use measurement statistics to get information about the state measured, motivates a generalization of the notion of 'measurement' to include procedures such as the one above: we may describe possible measurement outcomes by one, two, or more operators $M_c$ (which are in fact 'Kraus operators', objects associated to CPTP maps), with outcomes described by a slightly generalized Born rule</p>

<p>$$ \Pr\limits_{|\psi_0\rangle}(\text{result}=c) \;=\; \langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle \;, $$</p>

<p>where $M_c$ is a Kraus operator associated with your measurement, and with an update rule given by</p>

<p>$$ |\psi_1\rangle \;=\; \frac{M_c |\psi_0\rangle}{\sqrt{\langle \psi_0 | M_c^\dagger M_c | \psi_0 \rangle}} \;.$$</p>

<p>In order for the probabilities to be conserved (so that with certainty <em>at least one</em> of the measurement results occurs), we require $\sum_c M_c^\dagger M_c = I$. This is the more general form in your question, described by Nielsen and Chaung. (Again, this looks slightly better when describing states by density operators.)</p>

<h3>General remarks.</h3>

<p>In general, any time that we introduce an ancilla (or collection of ancillas) <em>A</em>, interact a qubit (or register of several qubits) <em>X</em> unitarily with <em>A</em>, and then perform a projective measurement on <em>A</em>, this gives rise to a sort of measurement of <em>X</em>; the measurement operators can then be described by some collection of positive-semidefinite operators $M_c$ such that $\sum_c M_c^\dagger M_c \;=\; I$ (again so that probability is conserved).</p>

<p>The more general, weaker measurements described here are more closely related to POVMs, which allow you to easily describe measurement probabilities 'abstractly', without an explicit choice of transformations $M_c$, by providing operators $E_c = M_c^\dagger M_c$ and allowing you to use these in the Born rule to compute probabilities. As I alluded to both above and in my previous response, POVMs can be regarded as describing statistically-available information about a system.</p>

<p>Thinking of measurements in terms of Kraus operators (and in terms of a 'measurement result register' <em>A</em> as above) in this way allows you to subsume the notion of measurement into that of a CPTP map, which is an idea that I enjoy. (However, this doesn't really change things from an analytical standpoint, and isn't something you should worry about if you're not yet comfortable with CPTP maps).</p>

CC BY-SA 2.5
2010-09-10T14:18:10.433Z
2010-09-10T19:22:04.140Z
2010-09-10T19:22:04.140Z
Niel de Beaudrap
<p>First of all, why are observables represented by operators? In classical mechanics an observable is a real-valued function on phase space. It extracts information about values like the energy or momentum from the system but does not affect or interfere with it. If the observer is part of the system then measurement is a physical process and can change the system's evolution. For finite, non-infinitesimal time evolution to be unitary (i.e. preserve total probability) the infinitesimal time evolution must be Hermitian. This is Stone's theorem; it explains why operators in quantum mechanics are Hermitian.</p>

<p>If that makes sense, the formula $M\mid\psi\rangle / \sqrt{\langle\psi\mid M^\dagger M\mid\psi\rangle}$ follows from two things: </p>

<ul>
<li>$M$ describes infinitesimal time evolution of the measurement process for the observable. The successor to $\mid\psi\rangle$ is $M\mid\psi\rangle$ and by duality the successor to $\langle\psi\mid$ is $\langle\psi\mid M^\dagger$. </li>
<li>The norm $\langle\psi\mid\psi\rangle$ is the total probability of the state. Combined with the previous point, this shows the successor's total probability is $\langle\psi\mid M^\dagger \ M\mid\psi\rangle$. Dividing by the square root normalizes the state.</li>
</ul>

CC BY-SA 2.5
2010-09-28T07:52:43.290Z
Per, I'm not sure that the first bullet point is terribly clear. The $M$ in this case is one of a set of operators which make up a general measurement (presumably a POVM), and so the evolution is not deterministic. It is also not continuous, so the comment about infinitesimal evolution may be a little misleading. These are really conditional jumps.
Joe Fitzsimons
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2010-09-27T03:44:44.413Z
2010-09-28T05:46:28.257Z
2010-09-28T05:46:28.257Z
Per Vognsen
<p>I have just started (independent) learning about quantum computation in general from Nielsen-Chuang book.</p>

<p>I wanted to ask if anyone could try finding time to help me with whats going on with the measurement postulate of quantum mechanics. I mean, I am not trying to question the postulate; its just that I do not get how the value of the state  of the system after measurement comes out to $M_m/\sqrt{ &lt;\psi|M_m^+ M_m|\psi> }$.</p>

<p>Even though its just what the postulate seems to say, I find it really awkward that why is it this expression. I do not know if what I ask here makes sense, but this is proving to be something which for some reason seems to block me from reading any further,</p>

CC BY-SA 2.5
2010-09-09T02:48:41.240Z
The expression you've written, $M_m/\sqrt{ <\psi|M_m^+ M_m|\psi> }$, is not a state at all. I guess you meant to add a $|\psi>$ after that?
Robin Kothari
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2010-09-09T11:28:48.350Z
Yes, thats right. I meant to add an $|\psi>$ after that
Akash Kumar
CC BY-SA 2.5
2010-09-12T10:02:59.713Z
Please edit your question if you notice mistakes.
Jukka Suomela
<quantum-computing><quantum-information>
Quantum Computation - Postulates of QM
Akash Kumar
# Quantum Computation - Postulates of QM

(Asked by: username_0 on 2010-09-09)

I have just started (independent) learning about quantum computation in general from Nielsen-Chuang book.

I wanted to ask if anyone could try finding time to help me with whats going on with the measurement postulate of quantum mechanics. I mean, I am not trying to question the postulate; its just that I do not get how the value of the state of the system after measurement comes out to $M\_m/\sqrt{ <\psi|M\_m^+ M\_m|\psi> }$.

Even though its just what the postulate seems to say, I find it really awkward that why is it this expression. I do not know if what I ask here makes sense, but this is proving to be something which for some reason seems to block me from reading any further,

**Tags:** quantum-computing, quantum-information
> **Comments:**
> **username_1**: The expression you've written, $M_m/\sqrt{ <\psi|M_m^+ M_m|\psi> }$, is not a state at all. I guess you meant to add a $|\psi>$ after that?  
> **username_0**: Yes, thats right. I meant to add an $|\psi>$ after that  
> **username_2**: Please edit your question if you notice mistakes.  

---

## 5 answers

### Answer by username_3 on 2010-09-09. Score: 7 | Accepted answer

I don't know if this is an "explanation", but hopefully it is a useful "description".

More generally than projective measurements, one always measures an *operator*. (A projector is a special case of this.) So what does it mean to "measure an operator"?

Well, operators often correspond to 'observable' physical quantities. The most important in quantum mechanics, for instance, is energy; but one can also (sometimes indirectly) measure other quantities, such as angular momentum, *z*-components of magnetic fields, etc. What is being measured always gives real-valued results --- in principle, some definite result (e.g. an electron is in the 'spin +1/2' state as opposed to 'spin −1/2', or in the first excited energy level as opposed to the ground-state in a hydrogen atom, etc.), albeit each *a priori possible* result is realized with some probability.

We assign each of the real-valued outcomes of a measurement to a subspace. The way we do this is to describe a Hermitian operator --- *i.e.* an operator which associates a real eigenvalue to different subspaces, with the subspaces summing up to the whole Hilbert space. A projector is such an operator, where the real values are 0 and 1; *i.e.* describing that a vector belongs to a designated subspace (yielding a value of 1), or its orthocomplement (yielding a value of 0). These Hermitian operators are **observables**, and the eigenspaces are those for which the observable has a "definite" value.

But what about those vectors which are not eigenvectors, and do not have "definite" values for these observables? Here is the non-explaining part of the description: we project them into one of the eigenspaces, to obtain an eigenvector with a well-defined value. Which projection we apply is determined at random. The probability distribution is given by the familiar Born rule:

$$ \Pr\limits\_{|\psi\rangle}\bigl( E = c \bigr) \;=\; \langle \psi | \Pi\_c | \psi \rangle \;, $$

where $\Pi\_c$ is the projector onto the *c*-eigenspace of an 'observable quantity' *E* (represented by a Hermitian operator $A = \sum\_c \; c \cdot \Pi\_c$). The post-measured state is *some* projection of the state $|\psi\rangle$ onto *some* eigenspace of the observable *A*. And so if $| \psi\_0 \rangle$ is the pre-measurement state, $| \psi\_1 \rangle$ is the post-measurement state, and $\Pi\_c$ is the 'actual result' measured (*i.e.* the eigenspace onto which the pre-measurement state was actually projected), we have the proportionality result

$$ | \psi\_1 \rangle \;\propto\; \Pi\_c | \psi\_0 \rangle $$

by the projection rule just described. This is why there is the projector in your formula.

In general, the vector $| \psi'\_1 \rangle = \Pi\_c | \psi\_0 \rangle$ is not a unit vector; because we wish to describe the post-measurement state by another unit vector, we must rescale it by

$$ \|\;|\psi'\_1\rangle\;\| = \sqrt{\langle \psi'\_1 | \psi'\_1 \rangle} = \sqrt{\langle \psi\_0 | \Pi\_c | \psi\_0 \rangle} \;,$$

which is the square-root of the probability with which the result would occur *a priori*. And so, we recover the formula in your question,

$$ | \psi\_1 \rangle \;=\; \frac{\Pi\_c | \psi\_0 \rangle}{ \sqrt{ \langle \psi\_0 | \Pi\_c | \psi\_0 \rangle }} \;.$$

(If this formula seems slightly clumsy, take heart that it looks and feels a little bit better if you represent quantum states by density operators.)

**Edited to add:** the above should not be construed as a description of POVMs. A "positive operator valued measurement" is better seen as describing the *expectation value* of various measurable observables *Ec* in a collection { *Ec* }*c* ∈ C .


---

### Answer by username_4 on 2010-09-10. Score: 5

If you haven't seen them already I highly recommend Scott Aaronson's lecture notes ["Quantum Computing Since Democritus"](http://www.scottaaronson.com/democritus/), especially [lecture 9](http://www.scottaaronson.com/democritus/lec9.html). They really helped me as a non-expert and I have tried to distill his presentation to the main points [here](http://rweba.livejournal.com/429528.html) and [here](http://rweba.livejournal.com/429744.html).

As far as your specific query I think it helps build intuition if you can calculate some simple example's using the Born Rule and see how the Measurement Postulate works in practice.

I found it easiest to think of as "The probability of measuring the ith outcome is the square of the amplitude of the ith element of the state vector - if you do a change of basis to to the eigenvectors of the Operator."

This also ties in neatly with the intuition that quantum mechanics is probability with complex numbers - since the squares of the amplitudes must sum up to 1.

As long as you're studying quantum computing you might also want to check out this discussion of [Shor's algorithm](http://www.scottaaronson.com/blog/?p=208).

> **Comments:**
> **username_0**: Thanks to you Mugizi...Scott Aaronson's lecture notes seem really nice.  

---

### Answer by username_5 on 2010-09-27. Score: 4

Niel de Beaudrap's answer regarding Kraus Operators was very good. With regard to the Nielsen and Chuang text book, this means that one should read Chapter 2, then Chapter 8, and *then* the intervening chapters.

Moreover, the Kraus operator representation has an infinitesimal limit called a Lindbladian operator; broadly speaking, Lindbladian operators are to Kraus operators what a Lie algebra is to a Lie group. Carlton Caves' on-line notes "Completely positive maps, positive maps, and the Lindblad form" cover much of this material.

The advantage of working exclusively with infinitesimal Lindbladian operators instead of Kraus operators is that the Lindbladians pullback naturally onto non-Hilbert quantum state-spaces; these include the tensor network state-spaces that are becoming ubiquitous in quantum chemistry and condensed matter physics; moreover pullback techniques are ubiquitous in string theory too.

There is at-present no textbook that develops this geometric, non-Hilbert description of quantum dynamics ... but there should be! Textbooks that (with the above references) in aggregate cover cover the main ideas are John Lee "Smooth Manifolds", Frenkel and Smit "Understanding Molecular Simulation: from Algorithms to Applications", and Kloeden and Platen "Numerical Solution of Stochastic Differential Equations."

It's true that this is a lot of reading ... and this is why geometric quantum dynamics is not taught at the undergraduate level. This is a pity, because it is all too easy for undergraduates to acquire the fixed notion that the state-space of quantum dynamical systems is a linear vector space, even though this is not true in most large-scale practical calculations.

As for the state-space that Nature uses: no one knows—the experimental evidence for local (tangent-space) quantum linearity is fairly strong, yet the evidence for global (Hilbert-space) quantum linearity is fairly weak. In particular, high-precision molecular beam quantum dynamical experiments—which many textbooks hold forth as evidence of quantum linearity—can be simulated with the required relative precision of ~1/2^{65} on low-dimension tensor network state-spaces, with near-perfect dynamical symplecticity replacing near-perfect dynamical linearity.

For the above reasons, perhaps 21st century students should not accept the textbooks of the 20th century entirely at face value. But really, what 21st century student would want it any other way?

The above is how quantum systems engineers have come to embrace a mathematical toolset that melds geometric and algebraic naturality, and applies generally to classical, quantum, and hybrid dynamical systems.

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*Edit addition:* As a test of the feasibility of a geometric approach to practical quantum simulation, our Quantum Systems Engineering (QSE) Group supplemented Charlie Slichter's classic textbook *Principles of Magnetic Resonance* with an enhanced version of Chapter 3 "[Magnetic Dipolar Broadening and Polarization Transport in Rigid Lattices](http://faculty.washington.edu/sidles/ENC_2011/index.html#Slichter)".

This geometric transcription points naturally to multiple open questions in geometric dynamics; see for example the MathOverflow question "[In quantum dynamical simulations, what is the symmetric (Riemannian) analog of a Poisson bracket?](https://mathoverflow.net/questions/66506/in-quantum-dynamical-simulations-what-is-the-symmetric-riemannian-analog-of-a)"

> **Comments:**
> **username_6**: I've seen you wave the flag for this approach all over the net. With a suggestive sentence or two, could you give an idea of how the state spaces you mention are non-linear? With geometric quantization you start with a manifold M as the classical phase space but the quantum state space is the Hilbert space L^2(M). That is, even if the classical geometry is highly non-linear the quantum geometry is still linear, though it is of course much bigger (it has infinite dimension and so on).  
> **username_6**: Sorry, I told a white lie. You actually have to look at L^2 over a line bundle on M. But the basic point remains.  
> **username_5**: Per, what you say is true of the classic (mainly Russian) school of "geometric quantization", in which one starts with a classical system and seeks a quantum generalization of it.   But exactly the <i>opposite</i>  happens in Ashtekar/Schilling models of "geometric quantum mechanics", in which the starting point is symplectic/Lindbladian dynamics on a K&auml;hler manifold.  
> **username_5**: Hmmm ... let's format it better!   Per, in the (mainly Russian) school of "geometric quantization" one starts with classical dynamics and seeks a quantum generalization of it.   The opposite move is seen in Ashtekar/Schilling models of "geometric quantum mechanics", in which the start is symplectic/Lindbladian dynamics on a Kahler state-space, following which one: (1) exhibits classical dynamics as a limit induced by Lindblad flow, and/or (2) pulls-back onto Hilbert space as a large-N (spectral) approximation.  In engineering, the latter two methods are commonly used, yet not commonly taught.  

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### Answer by username_3 on 2010-09-10. Score: 4

## Addendum.

After re-considering the form of your question (*e.g.* the M†M in the denominator --- as opposed for instance to a single operator M, which suffices for projectors) and reconsulting my copy of Nielsen and Chaung, here are some supplementary details not covered by my previous answer. (I'm posting this as a separate answer due to length, and because I feel that this is even less of an 'explanation' than my previous answer.)

Suppose that our only means of measuring a qubit *X* is indirect: by a 'weak' interaction with an ancilla *A*, followed by a measurement on *A*.
We would like to be able to talk about these as being in a sense a way of measuring *X*. How might we describe such a measurement in terms of *X* alone? Well: suppose we can easily prepare *A* in the initial state $|+\rangle \propto |0\rangle + |1\rangle$, and perform a controlled unitary of the following sort, with *X* as the control and *A* as the target:

$$ U \;=\; \left[\begin{matrix} \quad1\quad & 0 & 0 & 0 \\ 0 & \quad1\quad & 0 & 0 \\ 0 & 0 & \cos(\tfrac{\pi}{12}) & \sin(\tfrac{\pi}{12}) \\ 0 & 0 & -\sin(\tfrac{\pi}{12}) & \cos(\tfrac{\pi}{12}) \end{matrix}\right] $$

We then measure *A* in the standard basis (so that *A* now stores the measurement result). This transforms the state of *X* as follows:

$\begin{align\*} |\psi\_0\rangle\_X \;=&\; \alpha|0\rangle\_X + \beta|1\rangle\_X
\\\\\mapsto&\; \alpha |0\rangle\_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle\_A + \tfrac1{\sqrt 2}|1\rangle\_A\bigr) \quad+\quad \beta |1\rangle\_X \otimes \bigl(\tfrac1{\sqrt 2}|0\rangle\_A + \tfrac1{\sqrt 2}|1\rangle\_A\bigr)
\\\\\mapsto&\; \alpha |0\rangle\_X \otimes \bigl(\tfrac1{\sqrt 2} |0\rangle\_A + \tfrac1{\sqrt 2}|1\rangle\_A\bigr) \quad+\quad \beta |1\rangle\_X \otimes \bigl(\tfrac{\sqrt 3}2|0\rangle\_A + \tfrac12|1\rangle\_A\bigr)
\\\\=&\; \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle\_X + \tfrac{\sqrt3\beta}{2} |1\rangle\_X \bigr) \otimes |0\rangle\_A \quad+\quad \bigl( \tfrac{\alpha}{\sqrt 2} |0\rangle\_X + \tfrac{\beta}{2} |1\rangle\_X \bigr) \otimes |1\rangle\_A
\\\\\mapsto&\;
\begin{cases}
|\psi\_1\rangle\_X \otimes |0\rangle\_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle\_X + \tfrac{\sqrt 3\beta}{2}|1\rangle\_X\bigr) \otimes |0\rangle\_A & \;\text{for the result 0; or } \\
|\psi\_1\rangle\_X \otimes |1\rangle\_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}|0\rangle\_X + \tfrac{\beta}{2}|1\rangle\_X\bigr) \otimes |1\rangle\_A & \;\text{for the result 1.}
\end{cases}
\end{align\*}$

In the equations above, note that if the result of the measurement is *c*, the final state $|\psi\_1\rangle$ of *X* is proportional to $|\psi'\_1\rangle = M\_c |\psi\_0\rangle$, where we define

$$ M\_0 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{\sqrt 3}{2} |1\rangle\langle 1|\;,\qquad
M\_1 \;=\; \tfrac{1}{\sqrt 2} |0\rangle\langle 0| + \tfrac{1}{2} |1\rangle\langle 1|\;;$$

and we may verify that the probabilities with which we obtain the measurement results are in each case $\langle \psi'\_1 | \psi'\_1 \rangle \;=\; \langle \psi\_0 | M\_c^\dagger M\_c | \psi\_0 \rangle$.

This is very close to describing the transformation of *X* in the same way that we describe projective measurements. But is this any sort of measurement, meaningfully speaking? Well: if we can do statistics on the results of multiple iterations of this procedure, and if *X* is initially in the standard basis, we would notice that there is a bias in when we obtain the '0' result: we obtain it more often when *X* is initially in the state $|1\rangle$.
If we can sample enough times to distinguish whether the measurement results are distributed more like $(\frac12,\frac12)$ or $(\frac34,\frac14)$, we can determine with high probability whether the qubit is initially in the state $|0\rangle$ or the state $|1\rangle$.

The similarity of the probabilities-and-update formulae to those of projective measurement, and the fact that we can use measurement statistics to get information about the state measured, motivates a generalization of the notion of 'measurement' to include procedures such as the one above: we may describe possible measurement outcomes by one, two, or more operators $M\_c$ (which are in fact 'Kraus operators', objects associated to CPTP maps), with outcomes described by a slightly generalized Born rule

$$ \Pr\limits\_{|\psi\_0\rangle}(\text{result}=c) \;=\; \langle \psi\_0 | M\_c^\dagger M\_c | \psi\_0 \rangle \;, $$

where $M\_c$ is a Kraus operator associated with your measurement, and with an update rule given by

$$ |\psi\_1\rangle \;=\; \frac{M\_c |\psi\_0\rangle}{\sqrt{\langle \psi\_0 | M\_c^\dagger M\_c | \psi\_0 \rangle}} \;.$$

In order for the probabilities to be conserved (so that with certainty *at least one* of the measurement results occurs), we require $\sum\_c M\_c^\dagger M\_c = I$. This is the more general form in your question, described by Nielsen and Chaung. (Again, this looks slightly better when describing states by density operators.)

### General remarks.

In general, any time that we introduce an ancilla (or collection of ancillas) *A*, interact a qubit (or register of several qubits) *X* unitarily with *A*, and then perform a projective measurement on *A*, this gives rise to a sort of measurement of *X*; the measurement operators can then be described by some collection of positive-semidefinite operators $M\_c$ such that $\sum\_c M\_c^\dagger M\_c \;=\; I$ (again so that probability is conserved).

The more general, weaker measurements described here are more closely related to POVMs, which allow you to easily describe measurement probabilities 'abstractly', without an explicit choice of transformations $M\_c$, by providing operators $E\_c = M\_c^\dagger M\_c$ and allowing you to use these in the Born rule to compute probabilities. As I alluded to both above and in my previous response, POVMs can be regarded as describing statistically-available information about a system.

Thinking of measurements in terms of Kraus operators (and in terms of a 'measurement result register' *A* as above) in this way allows you to subsume the notion of measurement into that of a CPTP map, which is an idea that I enjoy. (However, this doesn't really change things from an analytical standpoint, and isn't something you should worry about if you're not yet comfortable with CPTP maps).


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### Answer by username_6 on 2010-09-27. Score: 3

First of all, why are observables represented by operators? In classical mechanics an observable is a real-valued function on phase space. It extracts information about values like the energy or momentum from the system but does not affect or interfere with it. If the observer is part of the system then measurement is a physical process and can change the system's evolution. For finite, non-infinitesimal time evolution to be unitary (i.e. preserve total probability) the infinitesimal time evolution must be Hermitian. This is Stone's theorem; it explains why operators in quantum mechanics are Hermitian.

If that makes sense, the formula $M\mid\psi\rangle / \sqrt{\langle\psi\mid M^\dagger M\mid\psi\rangle}$ follows from two things:

* $M$ describes infinitesimal time evolution of the measurement process for the observable. The successor to $\mid\psi\rangle$ is $M\mid\psi\rangle$ and by duality the successor to $\langle\psi\mid$ is $\langle\psi\mid M^\dagger$.
* The norm $\langle\psi\mid\psi\rangle$ is the total probability of the state. Combined with the previous point, this shows the successor's total probability is $\langle\psi\mid M^\dagger \ M\mid\psi\rangle$. Dividing by the square root normalizes the state.

> **Comments:**
> **username_7**: Per, I'm not sure that the first bullet point is terribly clear. The $M$ in this case is one of a set of operators which make up a general measurement (presumably a POVM), and so the evolution is not deterministic. It is also not continuous, so the comment about infinitesimal evolution may be a little misleading. These are really conditional jumps.  

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