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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in July 2018
A measurement operator is not a measurement.
In the context of discrete measurement (e.g. in quantum computing), the simplest form of observation is on a single basis state, which only gives you a yes-or-no answer.
When the measurement has a spectrum of mutually exclusive outcomes (orthonormal basis), each may give a yes-or-no outcome, so you can tell which one has happened.
Regardless of the number of possible outcomes, what you can measure is the probability of each by repeating the experiment under the same condition.
For example, if you measure a polarised photon with a horizontal polariser ($\Ket H$), you will see a yes (pass) or no (block) outcome. The experiment does not tell you about the measurement on other basis state, like $\Ket V$, $\Ket L$ or $\Ket R$. You would need to measure on different angles and based on the probability map out the quantum state (polarisation of the photon).
To get the approximation of a quantum state, you would need to do many observations on the same basis state under the same condition to get a fair result of probability, then repeat the same measurement process on other basis states.
If the probabilities do not add up to unity, that means there are basis states missing in the measurement process. In short, each measurement only tells you about the basis states you are measuring and nothing else.
As quoted in the article Multi-Qubit Systems:
The measurement outcome space has therefore an associated direct sum decomposition of eigenspaces: $\boxed{V=S_1\oplus S_2\oplus\ldots\oplus S_k}$, where $k\le 2^n$ and represents the maximum number of possible outcomes the measuring device can give. Each outcome is in an eigenspace $S_i$.
A measurement operator is a mathematical device that maps a quantum state to another quantum state. You may interpret a measurement operator as an "exhaustive collection" of observations.
In discrete measurements, we often deal with operators in the form of $A=\sum_k\lambda_k\Ket k\Bra k$ where $\{\Ket k\}$ is an orthonormal basis and $\lambda_k\in\mathbb{R}.$ ($\lambda_k$ can be choosen to be unique to avoid degeneracy.)
An measurement operator is idempotent and Hermitian, but not unitary.
Since $\Ket k$s are orthonormal, $A\Ket k=\lambda_k\Ket k$. Therefore, $\lambda_k$ is the eigenvalue with eigenvector $\Ket k$, which is why $\{\Ket k\}$ is often called the eigenbasis.
As illustrated in the "Measurement" section, the setting of the appparatus determines the measurement operator, specially the eigenbasis.
The eigenvalue $\lambda_k$ may have physical meaning, but often in discrete measurement it is just a label which you can code messages with.
For example, if you code $\Ket0$ as +1 and $\Ket1$ as -1, a positive expectation value means tendency towards $\Ket0$ and negative towards $\Ket1$. Zero means a fair chance.
A density operator is in the form of $\rho=\sum_i p_i\Ket{\psi_i}\Bra{\psi_i}$, where $p_i\in[0,1]$ and $\Ket{\psi_i}$ is of unit length. It is Hermitian, and looks very much like a measurement operator.
The mathematical differences are that
However, the main difference between a density operator and a measurement operator is physical: a density operator represents either an ensemble in a mixed state or a pure state (or pure ensemble), while a measurement operator represents measurement outcomes as determined by the settings of the measuring apparatus.
The Trace of a density operator tells a lot about it.
Because the Trace of a matrix is the sum of its diagonol. For pure state $\Ket\psi\Bra\psi$, the diagonal consists of the square of each element in vector $\Ket\psi$, so $\Tr(\Ket\psi\Bra\psi)=\Verti\Ket\psi\Verti^2=1.$
For a mixed state density operator $\rho,~~\Tr(\rho) =\Tr(\sum_i p_i\Ket{\psi_i}\Bra{\psi_i}) =\sum_i p_i\Tr(\Ket{\psi_i}\Bra{\psi_i}) =\sum_i p_i=1 .~~ \boxed{\Tr(\rho)=1}$ for any density matrix $\rho$.
For a pure state density matrix, there is only one projector $\rho=P_1=\Ket{\psi_1}\Bra{\psi_1},~$ which is idempotent, so $\rho^2=\rho.~~ \boxed{\Tr(\rho^2)=1}$ for any pure state $\rho$.
| Name | \hspace{2em} Observable \hspace{2em} | Outcome |
|---|---|---|
| Projector | $P=\Ket k\Bra k$ | Yes or No |
| Measurement Operator | $A=\sum_k\lambda_k\Ket k\Bra k$ | A value of $\lambda_k$ |
The quantum system to measure can be a pure state or a mixed state.
$\begin{array}{ccccc} \text{State} & \text{Density Matrix} & \small\text{On Projector }P=\Ket k\Bra k & \small\text{On Observable }A=\sum_k\lambda_k\Ket k\Bra k & \small\text{Result Interpretation}\\ & & \small\text{(Yes or No Outcome)} & \small\text{(Expectation Value)} & \\\hline ~\\ \small\text{Pure} & \small\rho=\Ket\psi\Bra\psi & \small P\rho P=\Ket k\Braket{k\Verti\psi}\Braket{\psi\Verti k}\Bra k=qP, & \small\Braket{A}_\rho=\Braket{\psi\Verti A\Verti\psi}=\Tr(\rho A). & \small\text{Projection of }\Ket\psi\text{ on }\Ket k\text{ gives an amplitude,} \\ \small\text{State} & & \text{where probability }q=\Verti\Braket{k\Verti\psi}\Verti^2. & & \small\text{which squares to probability of "yes"} \\ ~\\ \small\text{Mixed} & \small\rho=\sum_i p_i\Ket{\psi_i}\Bra{\psi_i} & \small P\rho P=\sum_ip_i\Ket k\Braket{k\Verti\psi_i}\Braket{\psi_i\Verti k}\Bra k=qP, & \small\Braket{A}_\rho=\Tr(\rho A). & \small\text{An expectation value of outcome} \\ \small\text{State} & & \text{where probability }q=\sum_i p_i\Verti\Braket{k\Verti\psi_i}\Verti^2. & & \small\text{associated with the operator} \\ \end{array}$
Let us prove that in all cases above, the expectation value is always $\Tr(\rho O)$, where $O$ is the projector $P$ or observable $A$.
For a pure state $\Ket\psi$ projected to $\Ket k$ by $P$, its projection is $\Braket{k\Verti\psi}$.
$\Tr(\rho P) =\Tr\left((\Ket\psi\Bra\psi)(\Ket k\Bra k)\right) =\Tr\left(\Braket{k\Verti\psi}\Braket{\psi\Verti k}\right) =\Verti\Braket{k\Verti\psi}\Verti^2 ,$ which is the probability of a "yes" outcome. Taking the range of $[0,1]$, the probability is the expectation value of the projector $P$.
(Note: Trace is invariant under cyclic permutations.)
$\boxed{ \Braket{P}_\rho =\Tr(\rho P) }$ and $\boxed{ P\rho P =\Tr(\rho P)P }$ where $\rho$ is a pure state.
For a pure state $\Ket\psi$ measured by $A$, its projection on an arbitrary eigenbasis $\Ket k$ is $\Braket{k\Verti\psi}$. The probability is the square of such projection: $\Braket{\psi\Verti k}\Braket{k\Verti\psi}.$ The expectation value is the sum of the "coded messages" (the $\lambda$ values) over the probability distribution.
$\Braket{A}_\rho =\sum_k\lambda_k\left(\Braket{\psi\Verti k}\Braket{k\Verti\psi}\right) =\Bra\psi\left(\sum_k\lambda_k\Ket k\Bra k\right)\Ket\psi =\Bra\psi A\Ket\psi .$
$\Tr(\rho A) =\Tr\left(\Ket\psi\Bra\psi\sum_k\lambda_k\Ket k\Bra k\right) =\sum_k\lambda_k\Tr\left(\Braket{k\Verti\psi}\Braket{\psi\Verti k}\right) =\sum_k q_k\lambda_k =\Braket{A}_\rho,~$ where $q_k=\Verti\Braket{k\Verti\psi}\Verti^2$ representing the probability of $\Ket\psi$ projected on $\Ket k$.
$\boxed{ \Braket{A}_\rho =\Tr(\rho A) =\Bra\psi A\Ket\psi }$ where $\rho$ is a pure state.
For a mixed state $\sum_i\sqrt{p_i}\Ket{\psi_i}$ projected to $\Ket k$ by $P$, its projection is $\sum_i\sqrt{p_i}\Braket{k\Verti{\psi_i}}$.
$\Tr(\rho P) =\Tr\left(\left(\sum_i p_i\Ket{\psi_i}\Bra{\psi_i}\right)(\Ket k\Bra k)\right) =\sum_i p_i\Tr\left(\left(\Ket{\psi_i}\Bra{\psi_i}\right)(\Ket k\Bra k)\right) =\sum_i p_i\Tr\left(\Braket{k\Verti\psi_i}\Braket{\psi_i\Verti k}\right) =\sum_i p_i\Verti\Braket{k\Verti\psi_i}\Verti^2 ,$ which is the probability of a "yes" outcome. Taking the range of $[0,1]$, the probability is the expectation value of the projector $P$.
$\boxed{ \Braket{P}_\rho =\Tr(\rho P) }$ and $\boxed{ P\rho P =\Tr(\rho P)P }$ where $\rho$ is a mixed state.
For a mixed state $\sum_i\sqrt{p_i}\Ket{\psi_i}$ projected to $\Ket k$ by $A$, its projection is $\left(\sum_i\sqrt{p_i}\Ket{\psi_i}\right)\left(\sum_k\lambda_k\Ket k\Bra k\right).$
$\Tr(\rho A) =\Tr\left(\left(\sum_i p_i\Ket{\psi_i}\Bra{\psi_i}\right)\left(\sum_k\lambda_k\Ket k\Bra k\right)\right) =\sum_k\lambda_k\sum_i p_i\Tr\left(\Ket{\psi_i}\Bra{\psi_i}\Ket k\Bra k\right) =\sum_k\lambda_k\sum_i p_i\Tr\left(\Braket{k\Verti\psi_i}\Braket{\psi_i\Verti k}\right) =\sum_k\lambda_k\left(\sum_i p_i\Verti\Braket{k\Verti\psi_i}\Verti^2\right) .$
The last braket represents the probability of $\Ket{\psi_i}$ projected onto $\Ket k$ across the probability distribution $p_i$. It is the probability of obtaining a $\lambda_k$ outcome, so the expression is the expectation value of $\rho$ measured by $A$.
Note: Here we interpret $p_i$ as probability, assuming that the density matrix $\rho$ has been "normalised" with orthonormal basis. This is acceptable because for each mixed state "family" a "normalised" decomposition can be found by unitary transformation. (To be verified)
$\boxed{ \Braket{A}_\rho =\Tr(\rho A) }$ where $\rho$ is a mixed state.
KEPT
This property does not apply to mix states. Consider $\rho= p\Ket\psi\Bra\psi+(1-p)\Ket\phi\Bra\phi.~ \Tr(\rho)=p\Tr(\Ket\psi\Bra\psi)+(1-p)\Tr(\Ket\phi\Bra\phi)=1.$
However, $\rho^2 =p^2\Ket\psi\Bra\psi +p(1-p)\Ket\psi\Braket{\psi\Verti\phi}\Bra\phi +(1-p)p\Ket\phi\Braket{\phi\Verti\psi}\Bra\psi +(1-p)^2\Ket\phi\Bra\phi \le p^2+2p(1-p)+(1-p)^2=2p^2 .$
$\Tr(\rho)=p\Tr(\Ket\psi\Bra\psi)+(1-p)\Tr(\Ket\phi\Bra\phi)=1.$
$\rho^2 =\left(\sum_i p_i\Ket{\psi_i}\Bra{\psi_i}\right)^2 =\left(\sum_i p_i^2\Ket{\psi_i}\Bra{\psi_i}\right)+\ldots .$
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