What are Grassmann-Plucker relations?

In Duality, matroids, qubits, twistors and surreal numbers (recently submitted!) they

show that via the Grassmann-Plucker relations, the various apparent unrelated concepts, such as duality, matroids, qubits, twistors and surreal numbers are, in fact, deeply connected.

The paper includes many interesting topics which I am not well versed in (Grassmannian, Plucker Embedding, Hopf Map, Matroids, etc) & am wondering if it might be possible that someone could explain Grassmann-Plucker relations & how they are used in a quantum context?

• As far as I can remember, there is a work on the relationship between the geometrical structure of 3 qubits with twistor. You may find that paper. – XXDD Oct 29 '18 at 15:07
• @XXDD Is this the paper? I also found this (pdf). – meowzz Oct 29 '18 at 18:37
• @meowzz The one by Levay. For me it's interesting to find their geometric connection. – XXDD Oct 30 '18 at 1:50

(This answer is given from the point of view of the theory of quantization, in which quantum systems are described by means of a quantization map of a classical phase space into a quantum space. The Plucker embedding will be described as a special case of such a map. This case has many applications in quantum computation).

Classical dynamics takes place on manifolds. For example, the dynamics of a particle moving on a straight line is completely determined by its initial position $$x$$ and momentum $$p$$ (or equivalently velocity). The set of initial parameters needed to determine the dynamics, i.e., to solve the equations of motion is known as the system's phase space. In the above case it is $$\mathbb{R}^2$$. i.e., a two -dimensional vector space of all possible values of the position and the momentum. The dynamics is generated by functions on the phase called Hamiltonians $$H(p, q)$$ through the Hamilton's equation of motion:

$$\frac{dx}{dt} = \frac{\partial H}{\partial p}$$ $$\frac{dp}{dt} = -\frac{\partial H}{\partial x}$$

($$t$$ = time). The Hamiltonian corresponding to free motion is given by $$H(p, q) = \frac{1}{2m} p^2$$ Any reasonable function on the phase space can serve as a classical Hamiltonian. For example, the function: $$H(p, q) = p^2 + x^2$$ is the classical Hamiltonian of the Harmonic oscillator (which is the basis of continuous variable models.)

Phase spaces (i.e., sets of initial data) need not necessarily be vector spaces. This can happen, for example, if the particle's position is confined, for example, a particle in a box, or a particle with rotational degrees of freedom, in which the angular position is confined to lie on a sphere. In all the above cases, the particle momentum is not confined and can assume any value, thus the phase space has an infinite volume, as it has unbounded directions.

Quantum systems (including all quantum systems used in quantum computing, such as qubits, qudits, continuous variable models, toric codes, etc.) can be described by a classical system + a procedure of quantization, in which the phase space geometrical manifold is traded by a quantum system Hilbert space and the Hamiltonian functions are traded by operators on the Hilbert space. There is no unique procedure applicable to all kinds of systems, different quantization procedures often give slightly different results, but nevertheless, I'll describe one of these procedures which is certainly applicable at least when the quantum Hilbert spaces are finite dimensional (such as the case of qudits).

First, let me remark that a Hilbert space does not describe the quantum mechanical set of pure states because in quantum mechanics there is no relevance to the overall magnitude and phase of a state vector; the pure states are described by rays; thus, we are talking about a projective Hilbert space which is a Hilbert space with an equivalence relation:

$$|\Psi\rangle \sim c |\Psi\rangle, \quad c \in \mathbb{C}, \quad c \ne 0$$

When the Hilbert space is finite dimensional, the projective Hilbert spaces are called projective vector spaces or simple projective spaces, for example the projective vector space corresponding to an $$n$$ dimensional complex vector space is called a complex projective space and denoted by $$P(\mathbb{C}^n) \cong \mathbb{C}P^{n-1}$$ (Its dimension is $$n-1$$, one dimension less due to the equivalence relation).

The quantization procedure in this case reduces to an embedding of a classical phase $$M$$ space into a quantum space $$Q$$ of states which an appropriate projective vector space: $$M \overset{i}{\rightarrow} Q = \mathbb{C}P^{n-1}$$ In each quantization method, there is a recipe of how given a classical Hamiltonian function, one can construct a corresponding quantum operator (at least for a certain class of functions).

When the Hilbert spaces are finite dimensional, such as in the qudit case, the corresponding phase spaces have finite volume.

One of the most amazing things in the above quantization procedure is that in the case of a qudit, the complex projective space is also the classical phase space. Please see Ashtekar and Schilling.

This does not mean that quantum mechanics is equivalent to classical mechanics. It only means that the space of classical pure states is the same as the space of quantum pure states. The difference lies in the process of measurement.

Let me remark that the qudit is a representative case where the dimension of the quantum Hilbert space is finite, in this case the volume of the classical phase space is also finite. This is a general principle.

The above complete correspondence breaks in cases other than a single qudit. For example, for a set of two $$n$$-dimensional qudits, the phase space $$M = \mathbb{C}P^{n-1} \otimes \mathbb{C}P^{n-1}$$ while the quantum space is $$Q = \mathbb{C}P^{2n-1}$$. The quantization map: $$M \overset{i}{\rightarrow} Q$$ in this case is a special case of the Segre embedding mentioned in AHusain's answer.

Another case of with a finite dimensional Hilbert case is the case of fermions. A set of $$k$$ fermions living in an $$n \ge k$$ dimensional Hilbert space can assume only certain entangled state vectors which are fully antisymmetric (because fermions cannot be in the same state), for example a set of two fermions ($$k=2$$) on a $$n=4$$ dimensional Vector space can assume only the following state vectors

$$|\Psi\rangle = c_{12} v_1\wedge v_2 + c_{13} v_1\wedge v_3 + c_{14} v_1\wedge v_4 + c_{23} v_2\wedge v_3 + c_{24} v_2\wedge v_4 + c_{34} v_3\wedge v_4$$

(The wedge $$\wedge$$ is the antisymmetric tensor product: $$v_i \wedge v_j = v_i \otimes v_j - v_j \otimes v_i$$)

The complex dimension of this vector space is $$6$$ and of the corresponding projective vector space is $$5$$ (the real dimension is 10).

The classical phase space of the above set of fermions can be obtained as follows: Taking a fixed fermionic state, for example:

$$|\Psi\rangle = v_1\wedge v_2$$

The vectors $$v_i$$ are 4 dimensional; the phase space is the orbit of the action of the unitary group $$U(4)$$ on this fixed vector:

$$g \cdot |\Psi\rangle = gv_1\wedge gv_2, \quad g \in U(4)$$

Now, if $$g$$ acts only within the two dimensional subspace spanned by $$v_3$$ and $$v_4$$, it clearly does not change the fermion state; also if $$g$$ acts only within the two dimensional subspace spanned by $$v_1$$ and $$v_2$$, it also does not the fermion state because it only changes the basis, thus there is a subgroup $$U(2) \times U(2)$$ which does not change the initial state, thus the phase space in this case is given by:

$$Gr(2, 4) = \frac{U(4)}{U(2) \times U(2)}$$

This manifold is called the complex Grassmann manifold. The dimension in our case is: $$4^2-2^2-2^2 = 8$$. The quantization map, i.e., the embedding:

$$Gr(2, 4) \overset{i}{\rightarrow} \mathbb{C}P^{5}$$

is called the Plucker embedding (this term is applicable in the general case, for arbitrary $$k$$ and $$n$$ ) . It is clear from comparing the dimensions ($$8 < 10$$) that not every state in the projective Hilbert state can be obtained from a point of the Grassmannian, i.e., from a unitary rotation of a fixed initial state. Thus, if we take a general element in the projective space $$\mathbb{C}P^{5}$$, there will be certain relations that it must satisfy to be a unitary rotation of a fixed elements, these are called the "Plucker relations"

In our example there is a single Plucker relation: $$c_{12} c_{34} -c_{13} c_{24} + c_{14} c_{23} = 0$$ (These relations are necessarily homogeneous because both manifolds are projective. Please see for example the following article by Smirnov, where the Plucker embedding is explained in some detail, the above equation appears in example 2.11).

One use of the Grassmann manifold is in the solution the Schrödinger equation for fermions. Instead of looking for the ground state in the entire Hilbert space, we can formulate a variational problem running only on vectors belonging to the Grassmannian. This procedure, known as the Hartree-Fock method, results in an approximate solution. (This point was also mentioned in AHusain's answer.

Please see the following article by Karle and Pachos analyzing the geometry of the Grassmannian $$Gr(2,4)$$ from the holonomic quantum computation point of view.

The Grassmann manifold appears also as the ground state manifold of stabilizer codes, please see for example the following article by Zheng and Brun.

• Thank you for such an insightful answer! In the paper they mention anyons & I was wondering if you might be able to potentially comment? – meowzz Oct 29 '18 at 18:43
• Apart from their exceptional braiding properties (leading to their potential use in topological quantum computation), anyons being particles which live on two-dimensional planes can have fractional or even irrational spin. (In two spatial dimensions there is no restriction to integer or half integer spins as in our three- dimensional world). This is the motivation of the author. I found in an earlier article of his: arxiv.org/abs/1611.09699v1 two references that he gives of spinning particles of irregular spins appear: – David Bar Moshe Oct 31 '18 at 9:16
• cont. arxiv.org/abs/1008.2334 Mezincescu and Townsend where spin $\frac{1}{4}$ particles appear as excitations of supersymmetric strings in 2+1 dimensions; and inis.iaea.org/collection/NCLCollectionStore/_Public/23/068/… where particles with the same spin $\frac{1}{4}$ are obtained as representations of the group SL(2, $\mathbb{R}$). Beyond these two references, a lot of work has been done on spinning anyons. – David Bar Moshe Oct 31 '18 at 9:16
• Thank you! I am particulary interested in anyons as the implication from the paper I linked is that they relate to transfinite numbers. In addition, the concept of infinite numbers of spin (pg. 13) is something I would like to further understand. I could potentially ask as a new question? – meowzz Oct 31 '18 at 15:25
• From the earlier paper: "game theory may lead to even more interesting applications" This is what I have been working on! Unfortunately I have not been able to formulate it such a way that is concise & comprehensible.. – meowzz Oct 31 '18 at 15:34

Consider a single fermionic state given by a single Slater determinant. Let the Hilbert space for the single mode be $$\mathbb{C}^n$$. Let there be $$k$$ fermions. Then that single Slater determinant is the Plucker embedding with the same notation as on the wiki page. The task of the Plucker relations is then to identify this embedding from the full projectivized fermionic Hilbert space. What homogenous equations do you need to impose on the entries of the vector in order to know that it came from a single Slater determinant? That is the question that these relations answer.

Edit: I had left it as unsaid implication, but @DavidBarMoshe made a good point about being explicit about it. Working within this embedding is beginning premise of Hartree-Fock.

This is analogous to the similar question about the Segre embedding judging which states are entangled or not. That is the comment about this being a determinantal variety.

I am not familiar with the linked paper, so I have no comment about how they use these definitions.