Expectation of Single-Qubit Property in Two-Qubit System

Consider a two-qubit system in quantum state $\psi$ . This system may be entangled, so it may not be possible to “separate it out” into a tensor product of individual qubit states. The first qubit has a property given by operator $M$ ; what is the expectation of $M$ ?

My textbook (A Short Introduction to Quantum Information and Quantum Computation by Michel Le Bellac) says that the expectation is basically given by the expectation of the four-dimensional operator $M \otimes I$ (based on the two-qubit system state.) But I don’t actually think this is new physics; it should be derivable from just the problem formulation and the other axioms of quantum mechanics. So how can we do this?

First, for better clarity and generality, let’s say that instead of qubits, we are dealing with finite-dimensional quantum states. (This basically generalizes 2-dimensional to $n$ -dimensional. Infinite-dimensional quantum states, as we see for example with the position and momentum operators, can get more complicated and also involve concepts like the Dirac delta distribution for the position operator eigenfunctions, so we won’t discuss those here.) Thus, we have a system in state $\psi$ that is composed of two objects. Each object’s state belongs to an $n$ -dimensional Hilbert space. However, the objects may be entangled, so the composite state may not just be a tensor product of the individual states. We have an $n$ -dimensional operator $M$ that corresponds to a property of the first object. Given $\psi$ , what is the expectation of $M$ ?

Well, expectation definitionally is the sum of possible values of $M$ weighted by their probabilities. It is a postulate of quantum mechanics that the form of $M$ (which is given to us) must have eigenvectors $a_{1},\ldots,a_{n}$ that form a basis of the first object’s state space. Let this basis be denoted by $A = \left\{ a_{1},\ldots,a_{n} \right\}$ .

It is also a mathematical fact that if $V,W$ are two vector spaces with bases $B_{V},B_{W}$ respectively, then a basis of the tensor product space $V \otimes W$ is given by $B_{V} \otimes B_{W}$ . (Maybe with some additional technical conditions like that $V,W$ must be over the same field. Here, we’re anyway always working over the field $\mathbb{C}$ , so such technical conditions will always be satisfied.)

Thus, it is a fact that the two-object system state space has a basis of $A \otimes A$ . Therefore, every such state can be expressed as $\sum_{1 \leq i,j \leq n}^{}{c_{ij} \left( a_{i} \otimes a_{j} \right)}$ for complex coefficients $c_{ij}$ .

Now, the possible values of $M$ (which is only applicable to the first object) are given by its eigenvalues $\lambda_{i}$ , and say we label these $\lambda_{i}$ so that for each $i$ , $\lambda_{i}$ is associated with $a_{i}$ (assume non-degenerate spectrum for simplicity, I’m sure the logic is similar without this assumption too.) Then, we have

$\displaystyle E(M) = \sum_{i}^{}{\lambda_{i}P\left( \lambda_{i} \right)},$

and the probability of getting $\lambda_{i}$ is given by the Born Rule postulate, as equal to the probability that, when measured in the $A \otimes A$ basis, the two-object system has the first object in state $a_{i}$ . This is equal to

$\displaystyle \sum_{j}^{}\left| c_{ij} \right|^{2}.$

Thus,

$\displaystyle E(M) = \sum_{i}^{}\left( \lambda_{i}\sum_{j}^{}\left| c_{ij} \right|^{2} \right).$

Now, is this always equal to $E\left( M \otimes I_{n} \right)$ on $\psi$ ?

Let’s consider the special case of two qubits; this is the same as saying $n = 2$ . Let

$\displaystyle M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}.$

Then,

$\displaystyle M \otimes I_{2} = \begin{pmatrix} a\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} & b\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \\ c\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} & d\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \\ \end{pmatrix} = \begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \\ \end{pmatrix}.$

Let

$\displaystyle \psi = \sum_{1 \leq i,j \leq 2}^{}{c_{ij}\ \left( a_{i} \otimes a_{j} \right)},$

where

$\displaystyle Ma_{1} = \lambda_{1}a_{1},$

$\displaystyle Ma_{2} = \lambda_{2}a_{2}.$

Then,

$\displaystyle \left( M \otimes I_{2} \right)\psi = \begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \\ \end{pmatrix}\sum_{1 \leq i,j \leq 2}^{}{c_{ij}\ \left( a_{i} \otimes a_{j} \right)}.$

(It remains to finish this. Hopefully, we can see generalizable logic that convinces us of this equality for general $n$ .)

I also suspect that this situation can be generalized naturally to any subsystem of a “pure state” quantum system, not just one object in an $n$ -object system. Then, the common formulas that motivate the definition of the density operator are justified. In fact, the density operator can also be used to represent the state of any quantum subsystem, including a subsystem of a mixed state. Such a mixed state corresponds to a preparation of multiple pure-state systems, where each pure-state system has a state $\psi_{i}$ with probability $p_{i}$ . (Thus, mixed states may not be expressible as elements of a Hilbert space, which is the point of the distinction from pure states.)

Also, touching on a point made earlier, studying quantum computing has made me realize that it probably makes more sense when studying quantum mechanics to start off with qubits and other finite-dimensional systems first to get the essential quantum-mechanical axioms, then to talk about infinite-dimensional systems (like position wavefunctions) after that, where further math such as Schwarz distributions would be involved (since the eigenfunctions of the position operator are in fact translations of the Dirac delta distribution.) Thus, it would be best from a pedagogical perspective to study finite-dimensional quantum mechanics and Schwarz distributions as prerequisites to “full” non-relativistic quantum mechanics involving position, momentum, and similar observables.

First written sometime around September/October 2022, and updated with better understanding on December 2022.

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