# 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 in 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).

• 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 $|a\rangle \otimes |b\rangle$. 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 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?

• 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.
– user10851
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." Feb 24 '13 at 3:31
• That does seem reasonable
– user10851
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. Feb 24 '13 at 14:01
• I would argue that you've got the argument backwards: if the subsystems were always independent (i.e. unentangled), then the natural way to combine them would be to take the Cartesian product - which when endowed with the natural vector-space structure, becomes the direct sum rather than the tensor product, as user10851 pointed out. The fact that the subsystems aren't always independent, but can be entangled, is what necessitates using the tensor product rather than the direct sum. Aug 6 '17 at 7:47

No, it's not obvious at all. The (essentially identical) answers here and here provide some nice justification. The key idea is that if we only do measurements on a single subsystem, then the probabilities that come out of the Born rule do not change if we multiply the subsystem's state vector by a complex number. The tensor product is the only way to combine subsystems together into new state vectors which preserves this property once we consider the joint system.

I would like to add here some further theoretical content the the excellent answer by @joshphysics, since in my view this subject is not treated as it should deserve and there are several theoretical results on the subject which should be known.

Let us consider a quantum system $$S$$ described in the Hilbert space $$\cal H$$ and suppose that it is made of two independent parts $$S_1$$ and $$S_2$$. We want to discuss when this system can be represented in a suitable tensor product $$\cal H_1 \otimes \cal H_2$$, for some Hilbert spaces $$\cal H_i$$ associated with $$S_i$$, $$i=1,2$$.

In the rest of my answer I do not use the notion of tensor product as a priori description of independent subsystems, because I want to discuss when this description is feasible.

First of all the systems $$S$$, $$S_1$$, $$S_2$$ are pictured in terms of their observables. With a very great generality we can assume that these observables are bounded (the unbounded ones cam be obtained as limit in the strong operator topology from the bounded ones) and the sets of observables (including complex linear combinations of operators) are closed with respect to the product and the sum and the strong operator topology (this is necessary for implementing the standard spectral machinery).

This way we get a well-known structure called von Neumann algebra of observables. So there are three von Neumann algebras here: $$R$$ associated to $$S$$ and $$R_i$$ respectively associated to $$S_i$$, for $$i=1,2$$.

The selfadjoint operators $$A\in R$$ represent all of (bounded) observables of the system $$S$$. Similarly, The selfadjoint operators $$A_i\in R_i$$ represent all of (bounded) observables of the system $$S_i$$, $$i=1,2$$.

A typical situation is $$R = B(\cal H)$$, where $$B(\cal H)$$ denotes the full algebra of bounded operators $$A: \cal H \to \cal H$$ (if $$\cal H$$ is finite dimensional boundedness is automatic). However in case of presence of superselection rules or when there is a gauge group, not all selfadjoint operators over $$\cal H$$ represent observables, so the assumption $$R = B(\cal H)$$ is generally untenable.

Let us discuss how the notion of independent subsystems is presented in this picture. There are three requirements

1. $$R_i$$ are substructures of $$R$$: $$R_i \subset R$$, for $$i=1,2$$,
2. $$R_1$$ and $$R_2$$ are compatible: $$[A_1,A_2]=0$$ if $$A_1\in R_1$$ and $$A_2\in R_2$$,

3. we can independently assign states on $$R_1$$ and $$R_2$$ according to the requirement below called $$W^*$$-independence

[$$W^*$$-independence]. If $$T_1$$ and $$T_2$$ are statistical operators acting on the observables of $$R_1$$ and $$R_2$$ respectively ($$_{T_i} = tr(T_iA_i)$$), then there exists a statistical operator $$T$$ for the overall system (acting on $$R$$) such that $$tr(TA_i) = tr(T_iA_i) \quad \mbox{ for every A_i\in R_i, i=1,2.}$$

There are many implementations of the notion of independence in fixing states on subsystems and this is proper of the Hilbert space description of quantum theory.

Under the hypotheses (1)-(3) (also weakening them) it arises that the algebra generated by $$R_1$$ and $$R_2$$ (the finite sum of finite product so elements in the union of the algebras) is isomorphic to $$R_1\otimes R_2$$ in pure algebraic sense (without topological implications). This is however still quite far from the standard picture where also the Hilbert space is factorized $$\cal H = \cal H_1 \otimes \cal H_2$$ and the algebras $$R_1$$ and $$R_2$$ are interpreted as algebras of operators in the factors $$\cal H_1$$ and $$\cal H_2$$.

The converse is however true as I go to illustrate.

Suppose that $$\cal H = \cal H_1 \otimes \cal H_2$$ so that $$B(\cal H) = B(\cal H_1) \otimes B(\cal H_2)$$. Next we fix

• $$R=B(\cal H)$$,
• $$R_1= B(\cal H_1)\otimes I_2$$
• $$R_2= I_1 \otimes B(\cal H_1)$$

Above $$I_k$$ indicates the identity operator over $$\cal H_k$$. In particular $$A_1 \in R_1$$ takes the form $$A'_1\otimes I_2$$ for some $$A'_1 \in B(\cal H_1)$$ and an analogous fact is true for $$S_2$$.

In this case (1), (2) are satisfied trivially and (3) is true in an even stronger sense. If $$T_1$$ acts as a statistical operator over $$R_1$$ it can always be written as $$T'_1\otimes I_2$$, for some statistical operators in the space $$\cal H_1$$, the analogue is true for $$S_2$$. A state $$T$$ satisfying (3) is always $$T= T_1'\otimes T'_2$$. This state has a further property (immediately arising from the basic properties of the tensor product) $$tr(TA_1A_2)= tr(T_1A_1)tr(T_2A_2) \quad \mbox{ for every A_i\in R_i, i=1,2.}$$ This property, which is stronger than $$W^*$$-independence, is called statistical independence.

We have so far seen that the standard representation of independent subsystems based on the notion of tensor product is in agreement with the general requirements (1),(2),(3) valid in whole generality for independent subsystems.

The natural question is the converse one, whether or not the structure of independent subsystems (requirements (1)-(3)) is always implementable by means of the notion of tensor product.

The answer is negative since there physically fundamental systems where the notion of tensor product is inappropriate to describe independent subsystems. Perhaps the most important case is that of observables of a quantum field localized in two causally separated regions of the Minkowski spacetime. The associated von Neumann algebras satisfy (1), (2) and (3), but in general it is false that the algebra of observables generated by both regions (the overall system) is representable as a tensor product of von Neumann algebras over a corresponding tensor product of Hilbert spaces. (The tensor product can be used when a certain technical condition called split property is valid.)

Are there sufficient conditions assuring that (1),(2),(3) are implementable by the standard use of tensor product over a tensor product of Hilbert spaces?

There is an important result due to von Neumann which actually is valid in almost all situations of standard quantum mechanics (not QFT and thermodynamics of extended systems).

Suppose that

• (a) $$R= B(\cal H)$$,
• (b) $$R_1$$ is a type-I factor
• (c) $$R_2$$ is made of all operators in $$R$$ which commute with all operators in $$R_1$$.

(b) deserves some explanation. $$R_1$$ is a factor if it does not includes non-trivial operators which commute with all operators of $$R_1$$ (in other words there are no superselection rules). The requirement type-I is technical and means that $$R_1$$ is algebraically (not necessarily unitarily) isomorphic to some $$B(\cal K)$$ for some Hilbert space $$\cal K$$. This condition is always true if $$\cal H$$ is finite dimensional, and it is false in QFT where factors of type-III take place (and this is the reason for the above mentioned failure of the tensor product representation in QFT).

Under the hypotheses (a),(b), and (c), then (1),(2),(3) are valid and there exist a couple of Hilbert spaces $$\cal H_1$$, $$\cal H_2$$, a unitary operator $$U: \cal H \to \cal H_1 \otimes \cal H_2$$ such that $$UR_1U^{-1} = B({\cal H_1})\otimes I_2$$ and $$UR_2U^{-1} = I_1 \otimes B(\cal H_1)$$.

This is the most common situation where the notion of tensor product is the fundamental building block to describe independent subsystems.

I think it is pretty obvious. Correct me if my argument is wrong somewhere.

In the classical case, if you want to describe e.g. the x-positions of two particles, you have a two-dimensional phase space to show the possible states -- and two is the sum of one and one. But a quantum state space is very different -- every point in the $$x_1$$ "axis" is a basis vector of its own, and likewise for $$x_2$$ -- the state vectors we speak of are vectors in the Hilbert space, and can be shown as distributions mapped on this $$(x_1,x_2)$$ plane, representing them as superpositions of these basis vectors.

So it makes perfect sense that the dimension of the product space is the product of the dimensions and not the sum. The total number of points in the $$(x_1,x_2)$$ plane -- which is the dimension of this new Hilbert space -- is the product of the number of points on the $$x_1$$ axis and the $$x_2$$ axis.

It's clear that the probabilities are multiplicative. Given states $$|\phi\rangle=\sum p(x)|x\rangle$$ and $$|\psi\rangle=\sum q(y)|y\rangle$$ in bases $$|x\rangle$$ and $$|y\rangle$$, it is clear that the magnitudes of the components of the state $$|\phi\rangle\otimes|\psi\rangle=\sum r(x,y)|x\rangle\otimes|y\rangle$$ where $$\otimes$$ (is the desired product representing composition) of the combined system are $$|r(x,y)|^2=|p(x)q(y)|^2$$. But -- as you ask in your question -- how do we know that $$r(x,y)=p(x)q(y)$$?

The idea is quite simple, though -- suppose we have a state like

$$\left( {\frac{1}{{\sqrt 2 }}\left| x \right\rangle + \frac{1}{{\sqrt 2 }}\left| y \right\rangle } \right) \otimes \left| z \right\rangle = \frac{u}{{\sqrt 2 }}\left| x \right\rangle \otimes \left| z \right\rangle + \frac{v}{{\sqrt 2 }}\left| y \right\rangle \otimes \left| z \right\rangle$$

Because we're representing two independent systems, we can just observe the first system, collapsing it to $$|x\rangle$$: then the combined state based on the left-hand-side is collapsed to $$|x\rangle\otimes|z\rangle$$. But based on the right-hand-side, this is $$u|x\rangle\otimes|z\rangle$$, and thus $$u=1$$ and similarly for $$v$$.

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.