In Nielsen and Chuang's text on Quantum Information and Computation, the measurement postulate is stated by using a collection of measurement operators and the outcomes are the indices of the measurement operators. See: http://books.google.com/books?id=65FqEKQOfP8C&lpg=PA87&ots=Pq9S_kl6GO&dq=measurement%20postulate&pg=PA84#v=onepage&q&f=false .

I'm little confused over the fact that the eigenvalue postulate for projective measurements (that the outcome of a projective measurement is one of the eigenvalues of the observable), as written in other quantum mechanics texts, does not emerge from the way the postulate is stated in Nielsen & Chuang. While the eigenvalues can serve the same purpose as the indices, they contain more information than just an index: don't they also carry scale information? That is, if one eigenvalue is twice another eigenvalue, then the measuring apparatus should read twice the value for one outcome as compared to the other. So, is this is a weaker form of the measurement postulate? Or are they equivalent and I'm missing something?

  • $\begingroup$ They use the eigenvalues for indexes. $\endgroup$ – MBN Mar 2 '11 at 14:20
  • $\begingroup$ @MBN, Please see the lines: "While the eigenvalues can serve the same purpose as the indices, they contain more information than just an index: don't they also carry scale information?...." in my question. $\endgroup$ – user2186 Mar 2 '11 at 14:41
  • $\begingroup$ I read it. What I am saying is that any complex number can be used as an index. The eigenvalues are complex (real in fact) numbers and can be used as indeces. I have not read the book, I am just guessing, but this is what is done in many places. $\endgroup$ – MBN Mar 2 '11 at 15:04

It is not a weaker form. The book quotes "index m refers to the measurement outcome". As Lubos wrote, the classical measurement outcome could be anything physical, which is not the focus of the postulate, so basically if $\mu(m)$ is a classical measurement outcome and $m$ is your corresponding index then it is implicit that $\mu$ is a mapping

$\mu :\{1,2,3,...N\} \rightarrow \mathbb{R} \times (\textbf{physical unit of observable}) $

  • $\begingroup$ That I understand. This was the reason I wrote the line: "eigenvalues can serve the same purpose as the indices". But the eigenvalues carry more information than just an index. That's why I mention the scale information too. So I understand that there is no motivation for quantum information theorists to look at the actual physical value of the outcomes as Lubos pointed out. But as far as quantum mechanics goes, the way the postulate is stated is not complete. That's all I wanted to clarify. $\endgroup$ – user2186 Mar 3 '11 at 9:24
  • $\begingroup$ I am not so sure about weak. The weaker form of Newton's third law extends its domain of applicability to magnetic forces as it relaxes the condition that forces act along the straight line along two particles. Here I see no relaxation, just that the actual eigenvalues will be replaced by representations of eigenvalues (indices) which map onto the real physical values because it suppresses unnecessary information. $\endgroup$ – Approximist Mar 3 '11 at 22:34

I have been looking at what is available of the book online, and believe that you are right that the Postulate 3 is (in a sense discussed below) weaker than the usual QM Projection Postulate.

Firstly there are some notational issues here. The $M_m$ are a family of operators, called Measurement Operators, indexed by $m$ which is a label for the outcome eigenvectors. However the eigenvalue equation itself does not play a part in Postulate 3, which is partly your point.

However the book does also discuss Projection Operators as a special case of Measurement Operators. Here $M = \Sigma m P_m$ where $M$ is a Hermitian operator and m is now the eigenvalue in its spectral decomposition into Projection operators $P_m$.

What is "special" about this special case is that $P_m$ is idempotent, physically that the "after measurement" state is as an eigenstate of $P_m$ now going to return the same result on immediate measurements. With a Measurement Operator there is the arbitrary state $M_m \Psi$ as after state, which might not be an eigenstate of $M_m$ (hence the notation could get a little confusing).

Later they introduce a Postulate 4 (on Tensor products and compound systems) and they claim to prove that:

Projective Measurement + Unitary (Postulate 2) + Postulate 4 "implement" General Measurement (Postulate 3)

So in a sense Postulate 3 is more general than Projective measurement as one could have non-projective measurements, but under all the other Postulates it all comes out the same in the end, if required.

I suppose that indeed part of the reason why this has all been done is because the eigenvalue does not do much more than label and distinguish the different outcomes in Quantum Computation and Quantum Information. Its actual physical value is less important in many calculations.

For example in the usual qubit situation we might have $|\Psi> = a |0> + b |1>$. Then the eigenvalue equation for $M_0$ might be $M_0 |0> =\alpha |0>$, but the $\alpha$ value is not of interest only its role in indicating that we now have a $|0>$ state.


Dear Siddharth, by measuring the projection operators $M_m$, one determines which eigenvalues the corresponding measured observables took. For example, the measurements distinguished by operators $M_m$ may correspond to the measurement of $x$, $p_y$, and $p_z$ of a particle. (I wanted to take a sufficiently complex example to make it clear that we may measure several observables at the same moment - and they don't have to be a set of the "most standard" observables that people usually talk about.)

So there is a one-to-one map between the projection operators and particular values of the three quantities: $$M_m |\psi\rangle = 1 |\psi \rangle \quad \Leftrightarrow \quad x |\psi \rangle = \lambda_{x,m} |\psi \rangle, \quad p_y |\psi \rangle = \lambda_{p_y,m} |\psi \rangle, \quad p_z |\psi \rangle = \lambda_{p_z,m} |\psi \rangle. $$ Of course, the quantities $x,p_y,p_z$ only take values that are the allowed eigenvalues.

Now, when we make a measurement, we assume that the laws describing the physical system are known. It also means that the precise map between the operators $M_m$ and the corresponding eigenvalues $\lambda_m$ of the three operators is known, too. The only thing that is unknown is the actual state of an actual physical system, controlled by these laws, at a given moment. That's what we measure. In this context, we don't try to "measure the laws of physics" or "all the allowed eigenvalues which are in principle possible". It's being assumed that we had known those general things in advance and that we have also chosen a convention what the operators $M_m$ physically represent - we have chosen the set of all the triplets of values $\lambda_m$.

You must understand that this is a textbook on quantum information so the precise values of the eigenvalues that represent the information - which projection operator $M_m$ had eigenvalue equal to one - is irrelevant. A quantum bit may be represented by a Hydrogen atom whose energy is $E=-13.6$ eV or $E=-13.6/4$ eV. But it may also be represented by the spin of the electron, $j_z=\pm 1/2$. The point of quantum information is that the two two-dimensional Hilbert spaces I just presented (and infinitely many other Hilbert spaces, labeled by different eigenvalues of different observables) are isomorphic. One is not interested in how the information is represented in the "hardware", just in the values (given by wave functions) of the quantum bits.

The amount of information you need to determined whether the Hydrogen atom was in the $n=1$ or $n=2$ state, and whether an electron was in $j_z=+1/2$ or $j_z=-1/2$, is the same: it's one qubit.

In classical computers, this would be analogous to the statement that we don't care whether bits are stored in the magnetic field on a hard disk or voltages in RAM, or by optical patterns on CD disks, or whatever else: we're only interested in the sequence of bits. Analogously, quantum information is the application of quantum mechanics where analogous technical questions are left to someone else: quantum information physicists only play with the quantum bits and their transformation, not with the hardware that is needed to realize such transformations in practice. But of course, as you correctly observe, every representation of a quantum bit in practice has to be linked to some Hermitean observable with some allowed eigenvalues - and they're usually not "zero" and "one".

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    $\begingroup$ Hi Luboš. I don't think there's anything in postulate 3 that requires $M_m$ to be projections. If we had some consistency principle like "if a measurement is immediately followed by an identical measurement, the same result will be obtained with probability 1", then I think it'd follow, but that's conspicuously missing from the formulation. (Unless idempotency (sans normalization) of $M_m$ follows from other things there, and I'm just too dumb to see it.) $\endgroup$ – Stan Liou Mar 2 '11 at 18:08
  • $\begingroup$ So am I right in saying that this is a weaker statement of the general quantum measurement postulate, since it leaves out the part about eigenvalues(which I understand is irrelevant for their purposes)? $\endgroup$ – user2186 Mar 2 '11 at 18:55

This is a nice question, with a nontrivial answer.

The reason why the rule where the eigenvalues do not figure is more general (and hence more widely applicable) is that one can label the outcomes of a measuring instrument with arbitrary values (e.g., by changing the scale from equidistant to nonequidistant), but it still measures the same stuff.

In fact, the statement Born's rule as usually stated is valid only when the spectrum is $\{0,1\}$ or $\{0,1,2,...\}$ and similar simple things, as in the basic textbook examples.

For example, measuring transitions from the ground state of a system is equivalent to measuring the eigenvalues of a Hamiltonian with ground state of zero energy. One never measures the exact eigenvalues (as the standard Born rule would suggest), as these are typically irrational numbers. But even a rough measurement of a spectral frequency tells you which transition was made, and hence give you the index of the eigenvalue/eigenvector.


Here we provide more details of the construction mentioned in Roy Simpson's answer. The question (v1) really has two parts:

  1. How are Hermitian observables $A:H \to H$ and projective measurements $P_m:H \to H$ related? This issue is discussed in many places, e.g. in this answer.

  2. How are projective measurements $P_m$ and general measurements $M_m$ related? This will be the main topic of this answer.

Let $H$ be the Hilbert space of the system. (We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.)

I) Projective measurement operators $P_m:H \to H$, $m\in I$, satisfy by definition

$$P_n P_m ~=~ \delta_{nm} P_m, \qquad P^{\dagger}_m~= P_m , \qquad \sum_{m\in I} P_m ~=~{\bf 1}_{H}. \qquad\qquad(1) $$

Here $I$ is an index set.

II) General measurement operators $M_m:H \to H$, $m\in I$, satisfy by definition

$$ \sum_{m\in I} M^{\dagger}_m M_m ~=~{\bf 1}_{H}.\qquad\qquad(2) $$

Consider a fixed $m\in I$. The probability $p(m)$ to measure outcome $m$ for the density operator $\rho:H\to H$ is

$$ p(m)~=~ {\rm tr}_H (M^{\dagger}_m M_m\rho).\qquad\qquad(3) $$

The collapse $\rho\longrightarrow \rho^{\prime}$ of the density operator, due to the general measurement, is

$$ \rho^{\prime}~=~ \frac{M_m\rho M^{\dagger}_m}{p(m)}.\qquad\qquad(4) $$ This is essentially Postulate 3 in Ref.1.

III) On one hand:

On the same Hilbert space $H$, the projective measurement operators (1) are a very special case of general measurement operators (2).

It is not hard to construct examples of general measurement operators that are not projective measurement operators (if the Hilbert space $H$ is fixed, cf. Section IV below). Note in particular, that if we repeat the general measurement $M_m$ with the same $m$, the doubly collapsed density operator

$$ \rho^{\prime\prime}~=~ \frac{M_m\rho^{\prime} M^{\dagger}_m}{{\rm tr}_H (M^{\dagger}_m M_m\rho^{\prime})}\qquad\qquad(5) $$

might in general be different from $\rho^{\prime}$. On the other hand, for projective measurement operators $\rho^{\prime\prime}=\rho^{\prime}$ always, mainly because of idempotency of $P_m$.

IV) On the other hand:

There is a way to realize general measurement operators $M_m:H \to H$, $m\in I$, which live on a Hilbert space $H$, as projective measurement operators $P_m:L \to L$, $m\in I$, on a larger Hilbert space $L:=H\otimes K$ by introducing a so-called ancilla Hilbert space $K~\cong~\mathbb{C}^{I}$ with an orthonormal basis $|m\rangle\in K$, $m\in I$, labeled by the same index set $I$.

This is explained in Section 2.2.8 of Ref. 1. The construction relies in particular on Postulate 4 for tensor products in Ref. 1. A sketched proof goes as follows.

  1. Choose a fixed normalized state $|a_0\rangle\in K$. Call the corresponding density operator $\rho_K:=|a_0\rangle\langle a_0|: K \to K$.

  2. Introduce an isomorphic copy $\tilde{H}$ of $H$ inside $L$ as $$H~\stackrel{\cong}\longrightarrow~ \tilde{H}~:=~H\otimes |a_0\rangle ~\subseteq ~H\otimes K~=:~L.\qquad\qquad(6) $$

  3. Define an isometry $U:\tilde{H} \to L$ as $$U~:=~ \sum_{m\in I} M_m \otimes |m\rangle \langle a_0 |.\qquad\qquad(7) $$ It is an isometry mainly because of eq. (2) and because $|m\rangle\in K$, $m\in I$, is an orthonormal basis.

  4. Extend the isometry $U:\tilde{H} \to L$ from eq. $(7)$ to a unitary operator $\tilde{U}:L \to L$.

  5. Define projective measurement operators $Q_m:L\to L$, $m\in I$, as $$Q_m~:=~ {\bf 1}_H\otimes|m\rangle \langle m|. \qquad\qquad(8) $$

  6. Define unitarily equivalent projective measurement operators $P_m:L\to L$, $m\in I$, as $$P_m~:=~\tilde{U}^{\dagger}Q_m\tilde{U} . \qquad\qquad(9) $$

  7. Recall the fixed density operator $\rho_K:=|a_0\rangle\langle a_0|: K\to K$ from above. Consider an arbitrary density operator $\rho_H: H\to H$. Define the product density operator $\rho_L:=\rho_H\otimes\rho_K: L\to L$.

  8. Consider for fixed $m\in I$ the projective measurement operator $P_m:L\to L$ defined in eq. $(9)$. Now apply Postulate 3 for projective measurement operator $P_m:L\to L$.

  9. The probability is $$p(m)~=~ {\rm tr}_L (P^{\dagger}_m P_m\rho_L)~=~\ldots ~=~{\rm tr}_H (M^{\dagger}_m M_m\rho_H).\qquad\qquad(10) $$

  10. The collapse $\rho_L\longrightarrow \rho^{\prime}_L$ of the density operator, due to the projective measurement, is $$ \rho^{\prime}_L~=~\frac{P_m\rho_L P^{\dagger}_m}{p(m)} ~=~\ldots~=~\rho^{\prime}_H~\otimes~\rho^{\prime}_K,\qquad\qquad(11) $$ where $$\rho^{\prime}_H~:=~\frac{M_m\rho_H M^{\dagger}_m}{p(m)}~=~{\rm tr}_K\rho^{\prime}_L,\qquad\qquad(12) $$ and $$\rho^{\prime}_K~:=~|m\rangle \langle m|~=~{\rm tr}_H\rho^{\prime}_L.\qquad\qquad(13) $$ Equations (10) and (12) reproduce Postulate 3 for the initially given general measurement operator $M_m:H\to H$. (The last equality of eqs. (12) and (13) uses partial trace.)


  1. M.A. Nielsen and I.L. Chuang, Quantum Computation and Quantum Information, 2011.

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