# Contradictory statements on product states for distinguishable particles in Quantum Mechanics

Page no. $$5$$ in Many-Body Theory Exposed! by Willem H Dickhoff & Dimitri Van Neck states the following:

The complex vector space, relevant for N particles, can be constructed as the direct product space of the corresponding state spaces [Messiah (1999)].

Whereas Wikipedia says :

the state space of a composite 2-(distinguishable) particle system is given by the tensor product

of the respective Hilbert spaces and hence the states, $$H_1\otimes H_2$$ and $$\big\{ |\psi_i \rangle \otimes |\psi_j \rangle \big\}$$, respectively.

The two statements seem contradictory (one says direct product while the other says tensor product). What am I missing?

• They are the same thing surely? – mike stone Jan 21 at 15:23
• @mikestone If you say direct product to a mathematician they will think you mean a direct sum, not a tensor product. – jacob1729 Jan 21 at 15:46
• I couldn't find a duplicate on physics, but the answere here on maths covers this appropriately. – jacob1729 Jan 21 at 15:57
• Can one show the occurance of "direct product space" in a mathematics text? – DanielC Jan 22 at 12:55
• Tensor product is the correct mathematical terminology in that particular context. I do not know of "direct product" in a particular mathematical context. We in physics use "direct product of Lie group/algebra representations", but this terminology must be traced to a pure mathematics textbook definition. – DanielC Jan 22 at 12:59

The direct product/sum$$^\dagger$$ and tensor product of two vector spaces $$V$$ and $$W$$ are not the same, though they look similar on the surface. The correct terminology is given by Wikipedia; we describe composite spaces as tensor product spaces, not direct product/sum spaces.

From a mathematical perspective, the difference is crucial. States in a direct product space are all so-called "product states", while the same is not true in tensor product spaces. This freedom gives rise to states which we call entangled in physics. However, from a terminology perspective, you will come across many physicists who say direct product when they mean tensor product. It's an unfortunate bit of linguistics, but what can you do?

The direct sum of $$V$$ and $$W$$ is denoted $$V\oplus W$$, and consists of all ordered pairs $$(v,w)$$ where $$v\in V$$ and $$w\in W$$, equipped with vector addition and scalar multiplication defined as follows:

$$(v_1,w_1) + (v_2,w_2) = (v_1+v_2,w_1+w_2)$$ $$\lambda (v,w) := (\lambda v, \lambda w)$$

Critically important is the fact that for any $$x\in V\oplus W$$, there must be some $$v\in V$$ and some $$w\in W$$ such that $$x=(v,w)$$.

The tensor product of $$V$$ and $$W$$ is denoted $$V\otimes W$$, and is really a different beast entirely. $$V\otimes W$$ consists of all ordered pairs $$(v,w)$$ and all linear combinations thereof, equipped with the following rules:

$$(v,w_1+w_2) := (v,w_1)+(v,w_2) \quad \text{and} \quad (v_1+v_2,w):=(v_1,w)+(v_2,w)$$ $$\lambda(v,w) := \underbrace{(\lambda v,w) = (v,\lambda w)}_{\text{defined to be equal to one another!}}$$

Let's take a quick, concrete look at how these spaces are different. Let $$V = \mathbb C^2$$ and $$W = \mathbb R^3$$, just for fun. The following operations correspond to the direct sum $$V\oplus W$$; for notational convenience, I will write $$(v,w)$$ as $$v\oplus w$$.

$$\pmatrix{i\\1}\oplus\pmatrix{2\\2\\3} + \pmatrix{0\\2i} \oplus \pmatrix{-1\\0\\1} = \pmatrix{i\\1+2i}\oplus\pmatrix{1\\2\\4}$$

$$6 \pmatrix{2i\\-1}\oplus\pmatrix{1\\-1\\2}= \pmatrix{12i\\-6}\oplus\pmatrix{6\\-6\\2}$$

Now we'll consider some similar operations with the tensor product space $$V\otimes W$$, where now I'll write $$(v,w)$$ as $$v\otimes w$$.

$$\pmatrix{i\\1}\otimes\pmatrix{1\\1\\2}+\pmatrix{i\\1}\otimes\pmatrix{-4\\2\\0}=\pmatrix{i\\1}\otimes\pmatrix{-3\\3\\2}$$ $$\pmatrix{0\\i}\otimes\pmatrix{1\\0\\0}+\pmatrix{i\\0}\otimes\pmatrix{0\\0\\1} \text{ cannot be combined or simplified!}$$ $$3 \pmatrix{1\\i}\otimes\pmatrix{2\\1\\0} = \pmatrix{3\\3i}\otimes\pmatrix{2\\1\\0} = \pmatrix{1\\i}\otimes\pmatrix{6\\3\\0}$$

My professor's notes say that for the direct sum, V and W should not have any common vector except the zero vector. $$\mathrm{dim}(V\oplus W)=\mathrm{dim}(V)+\mathrm{dim}(W)$$.

In what I showed above, I started with two vector spaces $$V$$ and $$W$$ and constructed $$V\oplus W$$, which consists of pairs $$(v,w)$$ equipped with pairwise addition and scalar multiplication. There is no condition whatsoever on $$V$$ and $$W$$, and in particular they can even be the same space.

However, let's go in the other direction. Let's say you have a big space $$V$$ and two subspaces $$W_1,W_2\subset V$$ such that for all $$v\in V$$, we can find unique $$w_1\in W_1$$ and $$w_2\in W_2$$ such that $$v=w_1+w_2$$. If that's true, then every pair $$(w_1,w_2)\in W_1\oplus W_2$$ corresponds to one and only one $$v\in V$$, given by $$v=w_1+w_2$$. Therefore, $$W_1\oplus W_2 \simeq V$$, where $$\simeq$$ means that they are isomorphic as vector spaces.

Now, $$V$$ and $$W_1\oplus W_2$$ are formally different spaces - the former consisting of individual vectors, and the latter consisting of pairs - which are in one-to-one correspondence. However, it's not too big of a leap to simply identify them with one another, and view $$(w_1,w_2)$$ and $$w_1+w_2$$ as being the same thing. Therefore, we informally drop the term "isomorphic" and simply say that $$V=W_1\oplus W_2$$.

So rather than starting from two vector spaces and constructing their direct sum, we started from one vector space and split it up into a direct sum of subspaces. Now, we can only do this if the splitting $$v=w_1+w_2$$ is always unique, and it's not too difficult to show that this requires that $$W_1\cap W_2=\{0\}$$ and nothing else. That's what your professors notes are referring to.

$$^\dagger$$For finite numbers of vector spaces, the direct product and direct sum are the same thing. For an infinite number vector spaces e.g. $$\bigoplus_{n=0}^\infty V_n := V_1\oplus V_2\oplus V_3 \oplus \ldots$$, these constructions differ slightly, but I'll ignore that for now.

• My professor's notes say that for the direct sum, $V$ and $W$ should not have any common vector except the zero vector. $\operatorname{dim}{(V \oplus W)} = \operatorname{dim}{(V)} + \operatorname{dim}{(W)}$. – Apoorv Potnis Jan 22 at 11:08
• @ApoorvPotnis I updated my answer to address your question. – J. Murray Jan 22 at 12:17
• I can't follow the math done in the tensor product example. Is there a typo on this line that says "cannot be combined or simplified!" ? – Steven Sagona Jan 22 at 17:41
• @StevenSagona There's not really any math to be done, I meant to give an example of an element of $V\otimes W$ which cannot be written as $v\otimes w$ for some $v\in V$ and $w\in W$. Such elements do not exist in the direct sum space $V\oplus W$. – J. Murray Jan 22 at 17:58
• Oh I see, they are three separate examples. – Steven Sagona Jan 22 at 18:01

Direct product, tensor product, Kronecker product and outer product mean approximately the same thing$$^*$$ - a product without summation over the inner indices. E.g., for two vectors: $$(\mathbf{a}\otimes\mathbf{b})_{ij} = a_i b_j$$ This is opposed to the inner product, dot product or scalar product, where $$\mathbf{a}\cdot\mathbf{b}=\sum_i a_i b_i$$

$$^*$$ There are nuances, depending on the context where the term is used and how the product is expressed. E.g., it does not make sense to talk about tensor product, when the quantities involved are not tensors. Kronecker product is usually applied when talking about block matrices, i.e., specific order of indexing and operating the quantities, etc.

• They only mean the same thing to physicists. To mathematicians, a direct product is a distinct operation, that for two vector spaces coincides with the direct sum. So for instance $R^3 \times R^3 \cong R^3 \oplus R^3 \cong R^6$ but $R^3 \otimes R^3 \cong R^9$. – jacob1729 Jan 21 at 15:49
• This is just wrong. I think you can get away with maybe saying Kronecker products, outer products, and tensor products are the same thing (they aren't exactly, as you note, but it's close enough to true that for a physicist it likely won't matter) but a direct product and tensor product are just two entirely different things. – Physical Mathematics Jan 22 at 1:05
• @Vadim Well, the two sources in the OP mean the same thing only insofar as the author of the book is using the wrong terminology, which is unfortunately not uncommon. It's true that notation varies, but at least in terms of vector spaces the differences between $\oplus$ and $\otimes$ are both standard and critically important. They are totally different constructions, and they are both in common usage in physics, so I don't view the distinction as technical. – J. Murray Jan 22 at 12:47
• @Vadim: $\mathbb R\times\mathbb R$ is $2$-dimensional as an $\mathbb R$ vector space. While $\mathbb R\otimes\mathbb R$ is one-dimensional as an $\mathbb R$-vector space. You should be the one explaining in which sense "they are the same". – Martin Argerami Jan 22 at 14:32
• @Vadim Martin is using $\times$ to mean direct sum/product and $\otimes$ to mean tensor product. $\mathbb R\times \mathbb R \simeq \mathbb R^2$, but $\mathbb R \otimes \mathbb R \simeq \mathbb R$. – J. Murray Jan 22 at 15:48

Unfortunately, the term "direct product (of vectors)" is commonly used in two completely different senses.

Some people use the term "direct product" of vectors to mean the same thing as the direct sum (up to some subtleties regarding infinite products). Under this definition, $$\mathbb{R}^n \times \mathbb{R}^m \cong \mathbb{R}^{n+n}$$.

Other people use the term "direct product" synonymously with "tensor product", so that $$\mathbb{R}^n \times \mathbb{R}^m$$ has $$nm$$, rather than $$n+m$$, degrees of freedom. In my experience, this usage is less common.

So there's a lot of confusion. Wikipedia is correct that you use the tensor product. Dickhoff and Van Neff are also correct under one definition of "direct product", but they are incorrect under the more common usage. Even Mathworld uses incompatible definitions: this page uses the first definition and this one uses (more or less) the second one.

That the direct product is the same as the direct sum for mathematicians is because they have changed a lot of mathematics terminology in recent years --- i.e since I was a student back in 1970's! ---due to the influence of category theory. Thus what used to be "orbits" of a group action are now called "coinvariants", and so on. What they mean is that "direct product" is their new terminology for the natural algebraic structure on the Cartesian product (set of pairs) of two vector spaces. That the Cartesian product, naturally equiped with a vector space structure, gives the direct sum is known even to physicists. It's just that we physicsists chose our terminology back in pre-categorical 1930's and have kept Cartesian product (plus structure) distinct from direct propduct.

• As a mathematician who is not into category theory, I find your answer a broad and mistaken generalization. In my neck of the woods (Functional Analysis/Operator Theory/Operator Algebras) the direct product and the Cartesian product are the same (and it is kind of unusual to use "Cartesian product" for infinite products). The structure depends on the structure of the elements in the product (a product of vector spaces is a vector space, a product of algebras is an algebra), simply defined entrywise. "Direct sum" is reserved for the subset of the product with finitely many nonzero entries. – Martin Argerami Jan 22 at 14:37
• @ Martin Argerami Thank you. I know what I wrote was a bit of a grumble about the language gap. And I do know that mathematicians (including mathematical physicists) do what they do to be careful. – mike stone Jan 22 at 14:53