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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
This page is a study note of "Quantum Computing: A Gentle Introduction"
by Eleanor Rieffel, Wolfgang Polak, The MIT Press © 2011.
A measurement is not a quantum gate. While a quantum gate changes the qubit state in a predictable way and is reversible, a measurement changes the state unpredictably and is irreversible. Measuring an $n$-qubit system (with $2^n$ dimensions) will "collapse" its state into one of the $n$-bit classical state (in its $n$ dimensional space). For a 3-qubit system, its 8-dimensional state space (superposition of $\Ket{000},\Ket{001},\ldots,\Ket{111}$, or $\Ket0,\Ket1,\ldots,\Ket7$) is measured to one of the 8 classical states (in a 3-dimensional space).
An example -- if a device measuring a 3-qubit system is aligned to the standard basis, $S_1$ has a single element basis $\{\Ket{000}\}$, and $S_2$ has $\{\Ket{001}\}$ etc, and $k=8$. The measurement outcome is entirely in one of the 8 subspaces $S_i$ where $i=1,2,\ldots,8$.
$S_i$ are orthogonal to each other, which means they are probabilistically mutually exclusive. The basis of $S_i$ does not have to be a standard basis. You may use, for example, the Bell States as measurement basis.
Generallly speaking, a measurement is a projection operation -- a "projector". It maps the state of the qubit system into an eigenstate of the projector. In other words, the measurement result is one of its eigenstates (and therefore projector dependent).
The dimension of the eigenspace $S_i$ equals to its degeneracy (the number of times its eigenvalues repeat). If the eigenvalue is unique, its corresponding eigenspace is of one-dimension. 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 $S_i$.
For example, for a 3-qubit system, $V=\{\Ket0,\Ket1,\Ket2,\Ket3,\Ket4,\Ket5,\Ket6,\Ket7\}$ if you measure all 3 qubits.
?? If you only measure the first qubit (left most), $V=\{\Ket0,\Ket1,\Ket2,\Ket3\}\oplus\{\Ket4,\Ket5,\Ket6,\Ket7\}$ and $k=2$.
The outcome space is a dual of the qubit system, so $\dim(V)=\sum_{i=1}^k\dim(S_i)=2^n$. In the degenerate case, $k<2^n$, and for some $S_i$ we have $\dim(S_i)>1$. i.e. outcome in $S_i$ is not unique. The measurement is leaving some "uncollapsed" freedom behind (as superposition of $\dim(S_i)$ number of orthonormal states). For example, measuring one qubit of a multi-qubit system will leave others "unmeasured", in a state space of $\dim(2^{n-1})$.
The action of a quantum gate acting on qubits is mathematically expressed as an operator acting on a quantum state.
The operator $\Ket{i}\Bra{j}$ will transform state $\Ket{j}$ to $\Ket{i}$ and everything else to null,
as $\big(\Ket{i}\Bra{j} \big)\sum_{k=1}^n a_k\Ket{k}=\Ket{i}\big(a_j\Braket{j\Verti j}\big)=a_j\Ket{i}$.
(Orthonormality is assumed in this context, so $\Braket{i\Verti j}=1$ if $i=j$ or 0 otherwise.)
Generally, $\begin{bmatrix} a & b \\ c & d \end{bmatrix} =a\Ket0\Bra0 +b\Ket0\Bra1 +c\Ket1\Bra0 +d\Ket1\Bra1 .$
The Pauli Gates: $X= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} =\Ket0\Bra1 +\Ket1\Bra0 ,~~ Y= \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} =-i\Ket0\Bra1 +i\Ket1\Bra0 ,~~ Z= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} =\Ket0\Bra0 -\Ket1\Bra1 .$
Let $~~ \Ket{\psi}=\alpha\Ket0+\beta\Ket1.~~ X\Ket{\psi}=\beta\Ket0+\alpha\Ket1.~~ Y\Ket{\psi}=-\beta i\Ket0+\alpha i\Ket1.~~ Z\Ket{\psi}=\alpha\Ket0-\beta\Ket1 .$
Do not mistaken, for example, $\Ket0\otimes\Bra1$ for $\Ket0\otimes\Ket1$. Their shorthands are $\Ket0\Bra1$ and $\Ket{01}$ respectively. $\Ket0\Bra1$ is a transformation of a single-qubit (from $\Ket1$ to $\Ket0$), while $\Ket{01}$ is a state of a 2-qubit system.
With matrices, $\Ket0$ is a 2x1 column vector and $\Bra1$ is a 1x2 row vector. $\Ket0\otimes\Bra1$ is a 2x2 matrix. $\Ket0\otimes\Ket1$ is a 4x1 column vector.
$\Ket0\otimes\Bra1 =\begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes [0,1] =\begin{bmatrix} 1\times[0,1]\\ 0\times[0,1] \end{bmatrix} =\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} .$
$\Ket0\otimes\Ket1= \begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix} =\begin{bmatrix} 1\times \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\ 0\times \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{bmatrix} =\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} =\Ket{01} .$
Extending this to a 2-qubit system with 4 dimensions: $\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$
$=a_{11}\Ket{00}\Bra{00} +a_{12}\Ket{00}\Bra{01} +a_{13}\Ket{00}\Bra{10} +a_{14}\Ket{00}\Bra{11} \\ +a_{21}\Ket{01}\Bra{00} +a_{22}\Ket{01}\Bra{01} +a_{23}\Ket{01}\Bra{10} +a_{24}\Ket{01}\Bra{11} \\ +a_{31}\Ket{10}\Bra{00} +a_{32}\Ket{10}\Bra{01} +a_{33}\Ket{10}\Bra{10} +a_{34}\Ket{10}\Bra{11} \\ +a_{41}\Ket{11}\Bra{00} +a_{42}\Ket{11}\Bra{01} +a_{43}\Ket{11}\Bra{10} +a_{44}\Ket{11}\Bra{11} .$
A general operator $\boxed{O=\sum_{i,~j}a_{ij}\Ket{i}\Bra{j} }$. When standard basis is used in a matrix representation, $O=[a_{ij}]$.
We can say $O$ "sends" $\Ket{j}$ to $a_{ij}\Ket{i}$. The verb might be "projects" or "measures".
Operator $O$ "measures" $\Ket{k}$ to $\sum_i a_{ik}\Ket{i}$ with probability $\sum_i\lvert a_{ik}\rvert ^2$, the sum of absolute square of the $k^{\text{th}}$ column.
So $\Braket{i\lvert O\rvert j} =\Bra{i} \left(\sum_p\sum_q a_{pq}\Ket{p}\Bra{q} \right)\Ket{j} =\Bra{i} \left(a_{ij}\Ket{i}\Bra{j} \right)\Ket{j} =a_{ij} .~~$ Sometimes we denote $a_{ij}$ as $O_{ij}$. i.e. $\boxed{\Braket{i\lvert O\rvert j}=O_{ij}~.}$
The action of a measurement is expressed as a projection of a quantum state vector onto an eigenspace of the "projector". Although mathematically a projector looks simular to an operator, the former changes the state unpredictably and irreversibly, while the latter is the opposite.
Let $V=S_1\oplus S_2\oplus\ldots\oplus S_k$, where $S_i$ are orthogonal. If we decompose the qubit state in this way: $\Ket{\psi}=\sum_{i=1}^k a_i\Ket{\psi_i}$, where $\Ket{\psi_i}\in S_i$ and $\lvert \Ket{\psi_i}\rvert =1$, The probability of measuring the eigenstate $\Ket{\psi_i}$ is $\lvert a_i\rvert ^2$.
For $\Ket{\psi_i}$ in a one-dimensional $S_i$, $\Ket{\psi_i}\Bra{\psi_i}$ projects $\Ket{\psi}\in V$ onto $\Ket{s_i}=\big(\Ket{\psi_i}\Bra{\psi_i}\big)\Ket{\psi}=\Ket{\psi_i}\cdot a_i\Braket{\psi_i\Verti\psi_i}=a_i\Ket{\psi_i}\in S_i.$
Note: $\Ket{s_i}$ is a projection and not necessarily a unit vector. In fact, $\lvert\Ket{s_i}\rvert=\lvert a_i\rvert$ where $a_i$ is the probability amplitude.
The projection operator, or projector, is $~~ P_i:V\rightarrow S_i~$, and $\boxed{P_i=\Ket{\psi_i}\Bra{\psi_i}.}~~ P_i\Ket{\psi}=a_i\Ket{\psi_i}.~~ \lvert P_i\Ket{\psi}\rvert =\lvert a_i\rvert .$
The projection measures the outcome with probability $\lvert P_i\Ket{\psi}\rvert ^2$. The normalised result state, or posterior state, is $\boxed{\Ket{\phi}=\frac{P_i\Ket{\psi}}{\lvert P_i\Ket{\psi}\rvert }}=\frac{a_i\Ket{\psi_i}}{\lvert a_i\rvert }$.
While the posterior state $\Ket{\phi}$ is a unity vector, the probability amplitude $a_i$ ($\le 1$ before the measurement) is the projection length of $\Ket{\psi}$ onto $\Ket{\phi}$.
For a 2-dimensional $S_i$ with basis $\{\Ket{\psi_{i1}},\Ket{\psi_{i2}}\},~~$ $\Ket{s_i}=\big(\Ket{\psi_{i1}}\Bra{\psi_{i1}}+\Ket{\psi_{i2}}\Bra{\psi_{i2}}\big)\Ket{\psi} =a_{i1}\Ket{\psi_{i1}}+a_{i2}\Ket{\psi_{i2}} .$
Generally, a projector from an $n$-dimensional $V$ onto an $s$-dimensional space $S$ with basis $\{\Ket{\alpha_i}\}$ is $\displaystyle P_s=\sum_{i=0}^{s-1}\Ket{\alpha_i}\Bra{\alpha_i}$.
The projection (or measurement result) is the sum of projections of $\Ket{\psi}$ onto individual basis vectors in $S$, an $n$ to $s$ dimensional projection.
Note: A projector only gives you a Yes or No answer in its corresponding subspace $S_i$ with a probability, and nothing about other subspaces. e.g. In a single-qubit, if you project $\Ket{\psi}=\frac{1}{\sqrt{2}}(\Ket0+\Ket1)$ onto subspace $\Ket0$, the projector $\Ket{00}\Bra{00}$ tells you that it is $\Ket0$ half of the times, and the rest it is not $\Ket0$, not that it is $\Ket1$. The subtle difference is that $P_s$ is projecting on $S$ with a probability, and does not tell the probability distribution of other subspaces.
Example:
In a 2-qubit system, the basis of $V$ is $\{\Ket{00}, \Ket{01}, \Ket{10}, \Ket{11}\}$.
The projector to measure only the first qubit along $\Ket0$ is $\Ket{00}\Bra{00}+\Ket{01}\Bra{01}$.
This projector sends any $\Ket{\psi}$ to the subspace with the left qubit being $\Ket0$.
i.e. $\{\Ket{00}, \Ket{01}\}.$
We can decompose $V=S_1\oplus S_2\oplus S_3$, where the basis of $S_1$ is $\{\Ket{00}, \Ket{01}\}$ (leaving $S_2~~\{\Ket{10}\}$ and $S_3 \{\Ket{11}\}$).
$P_1=\Ket{00}\Bra{00}+\Ket{01}\Bra{01}.~~$ Here are some measurement scenario:
If the system is in state $\Ket{\psi}=\Ket{00}$,
$P_1\Ket{\psi}=\Ket{00}$ with probability $\lvert P_1\Ket{\psi}\rvert ^2=1$. Same when measuring $\Ket{\psi}=\Ket{01}$.
In neither case, there is no "residue" probability for $S_2$ and $S_3$,
so we can tell that $\Ket{\psi}$ was not in either $S_2$ or $S_3$.
Likewise, when $\Ket{\psi}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{01})$,
$P_1\Ket{\psi}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{01})=\Ket{\psi}$.
The probability is unity, as $\Ket{\psi}\in S_1$ already.
No "residue" for $S_2$ and $S_3$.
This also implies that if the state has "collapsed" to a subspace, further projection on the same subspace does not alter it.
i.e. $P_1$ can be applied to $\Ket{\psi}$ multiple times with the same result of $\Ket{\psi}$.
${P_1}^n=P_1$.
However, state $\Ket{\psi}=\Ket{11}$ will measure $P_1\Ket{\psi}=0$ with zero probability. Same for $\Ket{\psi}=\Ket{10}$.
The "residue" probability is unity for $S_2\cup S_3$.
The measurement $P_1$ only confirms that the first qubit is not $\Ket0$
but gives no information about the second qubit at all.
(Information about the second qubit means the probability distribution of $\Ket{\psi}$ in $S_2\cup S_3$.)
When $\Ket{\psi}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{10})$,
$P_1\Ket{\psi}=\frac{1}{\sqrt{2}}\Ket{00}$ with probability $\lvert P_1\Ket{\psi}\rvert ^2=\frac{1}{2}$.
The result state $\Ket{\phi}=\frac{P_i\Ket{\psi}}{\lvert P_i\Ket{\psi}\rvert }=\frac{\frac{1}{\sqrt{2}}\Ket{00}}{\frac{1}{\sqrt{2}}}=\Ket{00}$.
Please note that while $\lvert \Ket{\phi}\rvert =\lvert \Ket{00}\rvert =1$, the certainty is after the measurement,
and this "collapse" happens only half of the times.
The "residue" probability for $S_2\cup S_3$ is half as well but there is no information about them.
The Bell State $\Ket{\psi}=\frac{1}{\sqrt{2}}(\Ket{00}+\Ket{11})$ is an interesting case. While measuring with $P_1$ would result in a 50% chance for $\Ket{00}$, the "residue" property is only in $S_3~~(\Ket{11})$, not $S_2\cup S_3$, but that is only if you knew beforehand that $\Ket{\psi}$ is the first or second Bell State.
A projector $P$ is self-adjoint ($P=P^\dagger$). Probability $\lvert P\Ket{\psi}\rvert ^2 =(\Bra{\psi} P^\dagger)~(P\Ket{\psi}) =\Bra{\psi}(P^\dagger P\Ket{\psi}) =\Braket{\psi\lvert(PP)\rvert \psi} =\Braket{\psi\lvert P\rvert \psi} .$
A Hermitian (aka Observerable) is also self-adjoint but is not necessarily a projector. e.g. A Hermitian can be a quantum gate. But Hermitian and measurement are closely related:
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