# Dimensions of a tensor product space

Susskind, in his 'Theoretical Minimum' book about QM says that given two 2-dimensional spaces, the product state of two vectors, one from each space, is determined by four real parameters. But, he later adds that the general state in the product space is determined by six real parameters.

P 165 Counting Parameters for the Product State Consider the number of parameters it takes to specify such a product state . Each factor requires two complex numbers $\alpha_u$ and $\alpha_d$ for Alice. $\beta _d$ and $\beta_d$ for Bob, which means we need four complex numbers altogether. That's equivalent to eight real parameters. But recall that the normalization conditions in Eqs. 6.4 reduce this by two. Furthermore, the overall phases of each state have no physical significance so the total number of real parameters is four. That's hardly surprising: it took two parameters to describe the state of a single spin, so two independent spins require four.

The most general vector in the composite set of states is:

$$\psi_{uu} | uu \rangle + \psi_{ud} | ud\rangle + \psi_{du} | du \rangle +\psi_{dd} | dd \rangle$$

P165/6 Again, we have four complex numbers but this time we only have one normalisation condition and only one overall phase to ignore. The result is that the most general state for a two-spin system has six real parameters. Evidently, the space of states is richer than just those product states that can be prepared independently Bob and Alice. Something new is going on. The new thing is called entanglement.

So my question is, what is the dimension of the product space? I thougt it was 4, but six degrees of freedom imply six dimensions, right?

• $\uparrow$ Which page? – Qmechanic Feb 24 '17 at 9:43
• @Qmechanic page 165 for the 4 parameter reference , page 166 for the six real parameters – user146020 Feb 24 '17 at 12:29
• He may be refering to general states in the tensor product space (6 real paremeters) vs. the subset of states of the form $|\psi\rangle \otimes |\phi\rangle$ (4 real parameters). – higgsss Feb 24 '17 at 12:50
• In general, the tensor product of a $p$-dimensional complex Hilbert space and a $q$ dimensional one has dimension $pq$. To get real dimension double the last number (the formula is clearly not valid for "the real dimensions"). Now of course not all of these parameters have physical meaning. In a one dimensional Hilbert space you have two real parameters and only one state altogether. The same is what he discusses above. How many real parameters do you need to specify a state, where vector $\neq$ state. Basically we are computing the real manifold dimension of the complex projective space. – Adomas Baliuka Feb 24 '17 at 13:37

The full Hilbert space of the states of the composite systems is $\mathbb{C}^2\otimes\mathbb{C}^2 \cong \mathbb{C}^4$, since you need to take arbitrary superpositions of the states $\lvert uu\rangle,\lvert ud\rangle,\lvert du\rangle,\lvert dd\rangle$. Since actual states are not encoded by vectors but by rays, we lose one complex degree of freedom by fixing the normalization and the phase of such a state. The actual space in which every point is a distinct state is the three-dimensional projective space $\mathbb{C}P^3$, described by six real parameters.
Likewise, the projective spaces of states of the subsystems with Hilbert spaces $\mathbb{C}^2$ are the one-dimensional projective spaces $\mathbb{C}P^1\cong S^2$, i.e. Bloch sphere. You need two real parameters to describe a point on the Bloch sphere - latitude and longitude.
The space of states of the composite system that consist of unique states of the subsystem is clearly $\mathbb{C}P^1\times\mathbb{C}P^1$, since we just pick a pair of states $(\lvert \psi\rangle_\text{Alice},\lvert \phi\rangle_\text{Bob}$). So we need four real parameters to describe such a pair, but six real parameters to descibe an arbitrary state of the composite system in $\mathbb{C}P^3$. The space of composite states that come from individual states of the subsystems is therefore smaller than the total space of allowed composite states, and the embedding $\mathbb{C}P^1\times\mathbb{C}P^1 \to \mathbb{C}P^3$ is called the Segre embedding. States not lying in the image of this embedding are called entangled, since you cannot reduce them to unambiguous pure states of the subsystems.