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What are boost and shift operators and why are they called so?

Quantum Computing Asked by Divy on September 5, 2021

In some texts I see $X$ and $Z$ Pauli operators as being said as boost and shift operators respectively.
But I came across some text that defines its own operators, namely:

$$
X vert jrangle = vert j+1,mod,drangle
Z vert jrangle = omega^jvert jrangle, quad omega = exp left( frac{2pi i}{d} right)
$$

I am confused as to what is the standard meaning of it, and why such a name.

Update: I realized Boost operator has something to do with quantum mechanics, where it is said to:

shift expectation value of momentum

Since I am not from physics background, it would be great if someone could explain it in simple words.

2 Answers

The shift operator takes his name from the fact that it shifts the position of its input, as in, it sends $1to2$, $2to3$ etc, with the last computational basis element being sent back to the first one: $dto 1$ (or the same thing starting with $0$, depending on notation).

As per the "boost" operator $Z$, I have usually seen those referred to as "clock matrices", as in the Wikipedia page. Such name comes from the fact that they are diagonal matrices, whose diagonal elements are phases of the form $omega_d^k$ where $omega_dequiv e^{2pi i/d}$. Represented in the complex plane, this are $d$ unit vectors pointing in equidistant directions, thus somewhat resembling hands on a clock.

Both definitions give the usual Pauli matrices for $d=2$.

As per their usefulness, it depends on the context. One thing that comes to mind is that they are used to find the "easy" examples of mutually unbiased bases, see e.g. here and here.

Answered by glS on September 5, 2021

The orthonormal basis $|jrangle$ of the $d$ dimensional finite Hilbert space corresponds to a configuration space of equally spaced clockwise ordered $d$ points on a circle $S^1$ or equivalently, the vertices of a $d$-dimensional regular polygon.

One may think of a point as a discrete location of a particle, then the shift operator $X$ shifts the particle position by one step clockwise. Thus, we may think of this orthonormal basis as the position basis.

More precisely, the configuration space may also be considered as the cyclic group $Z_d$, by defining the action of its points on themselves as modulo $d$ addition.

In ordinary quantum mechanics, where space is continuous, there is a dual basis: the momentum basis given by the action of the continuous Fourier transform on the position basis. In the discrete case, the momentum basis of the finite Hilbert space is given by the discrete Fourier transform of the position basis. $$|kranglerangle = sum_{j=0}^{d-1}omega^{-jk}|jrangle$$ It is not hard to see that the action of the operators on the dual basis is given by: $$X|kranglerangle = omega^{k} |kranglerangle$$ and: $$Z|kranglerangle = |k+1 mod dranglerangle$$ Thus, the operator $Z$, is the shift operator in the momentum basis. A shift in the momentum is a boost. This is the reason for this terminology.

So far, this is the standard explanation given in most quantum information resources. But it seems certainly strange at the first sight that a boost operator doesn't affect a position state (since a global phase shift doesn't define a new state). In elementary physics we are taught that a boost does indeed change the position of a particle. I'll try to explain this point and its manifestation in quantum information theory.

In classical mechanics, a (non-relativistic or Galilean) boost acts on the position and momentum observables as: $$qrightarrow q' = q+vt$$ $$prightarrow p' = p+mv$$ ($m$ is the particle mass, and $v$ is the boost velocity). While the second equation, shows that the Boost indeed introduces a jump in the value of the momentum, the first equation shows that it also changes the position.

However, the analog of a state in classical mechanics is not the position and momentum, but rather their initial values, i.e., at $t=0$. We see that indeed the Galilean boost introduces an instantaneous jump to the momentum but doesn't change the position at $t=0$ similar to the above discrete quantum boost, in agreement with the above finding.

In standard quantum mechanics (on a continuous space), the (infinitesimal) Galilean boost $mathbf{G}$ is given by: $$mathbf{G} = m mathbf{Q}$$ Where $mathbf{Q}$ is the position operator. This is clearly an operator which does not introduce a change in the particle position. This relation is given in equation (3.51) in Ballentine's book: Quantum mechanics: A modern development. Ballentine proves that the above relation is the only compatible choice for a system whose kinematics is governed by the Heisenberg commutation relation: $$[mathbf{Q}, mathbf{P}] = i hbar mathbf{I}$$ Returning to the discrete case, we observe that the momentum basis $|kranglerangle$ can also be modeled on a circle $S^1$ (or a regular polygon). The Cartesian product of the two discrete circles is called a phase space, and in our case, it is the collection of integer points on the torus $S^1times S^1$. Of circumference $d$. Equivalently, the phase space is the group $mathbb{Z}_d times mathbb{Z}_d$.

A phase space is the set of pure classical states. (In classical mechanics one can define a state with a definite discrete position and discrete momentum, but in the discrete Hilbert space there is no mutual eigenvector $X$ and $Z$). However, the phase space remains very important also in quantum theory, because both observables and states can be equally represented (in many ways) by functions on the phase space (in the case of operators, these functions are called operator symbols). In this representation, the associative composition of operators is called a star product which is useful for example in quantum tomography. In particular, Quasi-probability distributions describing quantum states are functions on the phase space. A prototypical example is the Wigner function, which is the Weyl symbol of the density matrix, whose expression for the discrete phase space:

$$W_{rho}(p, q) = frac{1}{d} mathrm{tr}(rho w(p, q)),quad p,qin mathbb{Z}_d,$$

where the Weyl operator is given by: $$w(p, q) = omega^{-frac{pq}{2}}Z^pX^q$$

Now, we know that boosts are elements of the Galilei group. Hence what is left is to see how the discrete Galilei group acts on the discrete phase and the corresponding Hilbert space.

The Galilean group consists of translations (shifts), rotations, boosts and time translations. When the configuration (position) space is a one-dimensional lattice $Z_d$ as in our case, there are no rotations. In addition, we do not want to commit to a single dynamics, so we consider the group without time translations (in non-relativistic theories this is possible, since the notion of simultaneity exists). This group is known by the name: isochronous Galilean group, which in our case consists of a single shift and a single boost. The action of its generators on the classical phase space: $$ x: qrightarrow q' = q+1mod d, quad prightarrow p'=p$$ $$ z: qrightarrow q' = q, quad prightarrow p'=p+1mod d$$ This action cannot be lifted to the quantum Hilbert space, where a central extension of the Galilean group is realized by means of the operators $X$ and $Z$. $$ZX=omega XZ$$ The origin of the central extension is the noncommutativity of the shift and boost after quantization. The need of a central extension is characteristic to quantization problems, where the action on the quantum space is realized by a central extension of the action on the phase space.

Thus, in the discrete phase space, the Galilean group is isomorphic to the (generalized) Pauli group generated by $X$, $Z$, $omega$.

For the action of the full Galilean group on the discrete phase space and on the finite Hilbert space, please see: ŠŤoviček and Tolar. For the definition of the generalized Pauli group on the finite (qudit) Hilbert space please see Tolar.

Answered by David Bar Moshe on September 5, 2021

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