# Should it be obvious that independent quantum states are composed by taking the tensor product?

My text introduces multi-quibt quantum states with the example of a state that can be "factored" into two (non-entangled) substates. It then goes on to suggest that it should be obvious1 that the joint state of two (non-entangled) substates should be the tensor product of the substates: that is, for example, that given a first qubit

$$\left|a\right\rangle = \alpha_1\left|0\right\rangle+\alpha_2\left|1\right\rangle$$

and a second qubit

$$\left|b\right\rangle = \beta_1\left|0\right\rangle+\beta_2\left|1\right\rangle$$

any non-entangled joint two-qubit state of $\left|a\right\rangle$ and $\left|b\right\rangle$ will be

$$\left|a\right\rangle\otimes\left|b\right\rangle = \alpha_1 \beta_1\left|00\right\rangle+\alpha_1\beta_2\left|01\right\rangle+\alpha_2\beta_1\left|10\right\rangle+\alpha_2\beta_2\left|11\right\rangle$$ but it isn't clear to me why this should be the case.

It seems to me there is some implicit understanding or interpretation of the coefficients $\alpha_i$ and $\beta_i$ that is used to arrive at this conclusion. It's clear enough why this should be true an a classical case, where the coefficients represent (where normalized, relative) abundance, so that the result follows from simple combinatorics. But what accounts for the assertion that this is true for a quantum system, in which (at least in my text, up to this point) coefficients only have this correspondence by analogy (and a perplexing analogy at that, since they can be complex and negative)?

Should it be obvious that independent quantum states are composed by taking the tensor product, or is some additional observation or definition (e.g. of the nature of the coefficients of quantum states) required?

1: See (bottom of p. 18) "so the state of the two qubits must be the product" (emphasis added).

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Consider an operator $A$ acting on the first factor and $B$ acting on the second factor. Since the factors are independent, we need to have that $(A \otimes B)(|a \otimes b\rangle = (A |a\rangle) \otimes (B|b\rangle)$ - do you agree? This fixes the state $|a \otimes b\rangle$ uniquely as $|a\rangle \otimes |b\rangle$. – Vibert Feb 23 '13 at 22:50
I saw a talk a few years ago in which the speaker, who was attempting to axoimize quantum mechanics, made a point of claiming that the tensor product rule should be regarded as a non-obvious axiom. The corresponding paper is here: arxiv.org/pdf/quant-ph/0405161v2.pdf – Rococo May 7 '15 at 22:21

Great question! I don't think there is anything obvious at play here.

In quantum mechanics, we assume that that state of any system is a normalized element of a Hilbert space $\mathcal H$. I'm going to limit the discussion to systems characterized by finite-dimensional Hilbert spaces for conceptual and mathematical simplicity.

Each observable quantity of the system is represented by a self-adjoint operator $\Omega$ whose eigenvalues $\omega_i$ are the values that one can obtain after performing a measurement of that observable. If a system is in the state $|\psi\rangle$, then when one performs a measurement on the system, the state of the system collapses to one of the eigenvectors $|\omega_i\rangle$ with probability $|\langle \omega_i|\psi\rangle|^2$.

The spectral theorem guarantees that the eigenvectors of each observable form an orthonormal basis for the Hilbert space, so each state $\psi$ can be written as $$|\psi\rangle = \sum_{i}\alpha_i|\omega_i\rangle$$ for some complex numbers $\alpha_i$ such that $\sum_i|\alpha_i|^2 = 1$. From the measurement rule above, it follows that the $|\alpha_i|^2$ represents the probability that upon measurement of the observable $\Omega$, the system will collapse to the state $|\omega_i\rangle$ after the measurement. Therefore, the numbers $\alpha_i$, although complex, do in this sense represent "relative abundance" as you put it. To make this interpretation sharp, you could think of a state $|\psi\rangle$ as an ensemble of $N$ identically prepared systems with the number of $N_i$ elements in the ensemble corresponding to the state $|\omega_i\rangle$ equaling $|\alpha_i|^2 N$.

Now suppose that we have two quantum systems on Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$ with observables $\Omega_1$ and $\Omega_2$ respectively. Then if we make a measurement on the combined system of both observables, then system 1 will collapse to some $|\omega_{1i}\rangle$ and system 2 will collapse to some state $|\omega_{2 j}\rangle$. It seems reasonable then to expect that the state of the combined system after the measurement could be any such pair. Moreover, the quantum superposition principle tells us that any complex linear combination of such pair states should also be a physically allowed state of the system. These considerations naturally lead us to use the tensor product $\mathcal H_1\otimes\mathcal H_2$ to describe composite system because it is the formalization of the idea that the combined Hilbert space should consists of all linear combinations of pairs of states in the constituent subsystems.

Is that the sort of motivation for using tensor products that you were looking for?

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Do you have any intuition as to why, e.g., we can't use $\mathcal{H}_1 \oplus \mathcal{H}_2$ to accomplish the same effect? I think this can be very subtle and confusing, especially once students start doing Clebsch-Gordan stuff and both methods of constructing combined spaces are in play. – Chris White Feb 24 '13 at 2:26
@ChrisWhite The best intuition I have regards dimension; tell me if what follows is convincing. The dimension of the tensor product is the product of the dimensions, and the dimension of the direct sum, is the sum. The idea behind a composite system is that the subsystems can in some sense "independently" simultaneously occupy their respective states. If system $1$ can occupy $n_1$ independent states, and if system $2$ can occupy $n_2$ independent states, then it seems reasonable to me that the composite should be able to occupy $n_1n_2$ independent states if the systems are "independent." – joshphysics Feb 24 '13 at 3:31
That does seem reasonable – Chris White Feb 24 '13 at 4:03
@joshphysics Since quantum states (or for that matter probability distributions) can be in a superposition, the dimensions indeed have to multiply (10 coins can be in $2^{10}$ states). However, to play devil's advocate, I will point out that, aside from the need for superpositions, it is in fact not obvious that the dimension multiply since in classical mechanics they add: one particle lives in a 6 dimensional phase space and ten particles live in a $6 \times 10$ dimensional phase space. – Dan Stahlke Feb 24 '13 at 14:01

Actually the story of the tensor product, which is typically told as an axiom in the world of quantum information, comes from the relativistic electron theory of Dirac.

In relativistic quantum mechanics one wavefunction is replaced by 4 wavefunctions, and a suitable contraction leads to a two-element array (1,0) or (0,1) which we call the spin.

In the case of 2 electrons the same theory leads to a 16-component wavefunction, and after some trimming we can get down to a 4 component object (called Pauli spinor for 2 electrons). This object and the various operations with the Hamiltonian are best described (in terms of algebra) by using tensor product of the states (1,0) and (0,1).

Obviously these tensor product correspond to spins, and since the idea of qubit comes from that of spin, the idea of tensor composition is now taken axiomatically in QIP.

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