The basic ideas and design implemented

Tensor product notation and the hideketlabel option

Entangled States and the Bell basis

Entangled States, Operators and Projectors

The basic ideas and design implemented

Suppose  and  are quantum operators and  are, respectively, their eigenkets. The following works since the introduction of the Physics package in Maple

In previous Maple releases, all quantum operators are supposed to act on the same Hillbert space. New: suppose that  and  act on different, disjointed, Hilbert spaces.

1) To represent that situation, a new keyword in Setup , , is introduced. With it you can indicate the quantum operators that act on a Hilbert space, say as in  with the meaning that the operator  acts on one Hilbert space while  acts on another one.

The Hilbert space thus has no particular name (as in 1, 2, 3 ...) and is instead identified by the operators that act on it. There can be one or more, and operators acting on one space can act on other spaces too. The disjointedspaces keyword is a synonym for hilbertspaces and hereafter all Hilbert spaces are assumed to be disjointed.

NOTE: noncommutative quantum operators acting on disjointed spaces commute between themselves, so after setting - for instance - , automatically,  become quantum operators satisfying (see comment (ii) on page 156 of ref.[1])

2) Product of Kets and Bras that belong to different Hilbert spaces, are understood as tensor products satisfying (see footnote on page 154 of ref. [1]):

while

3) All the operators of one Hilbert space act transparently over operators, Bras and Kets of other Hilbert spaces. For example

4) Every other quantum operator, set as such using Setup , and not indicated as acting on any particular Hilbert space, is assumed to act on all spaces.

5) Notation:

Tensor product notation and the hideketlabel option

According to the design section, set now two disjointed Hilbert spaces with operators  acting on one of them and  on the other one (you can think of  )

Consider a tensor product of Kets, each of which belongs to one of these different spaces, note the new notation using

You see that in the product of Bras, and also in the product of Kets, A comes first, then B.

Remark: some textbooks prefer a diadic style for sorting the operands in products of Bras and Kets that belong to different spaces, for example,  instead of the projector sorting style of  (2.3). Both reorderings of Kets and Bras are mathematically equal.

The display for (2.3) is now

Important: this new option only hides the label while displaying the Bra or Ket. The label, however, is still there, both in the input and in the output. One can "see" what is behind this new display using show, that works the same way as it does in the context of   CompactDisplay . The actual contents being displayed in (2.5) is thus (2.3)

Operators of each of these spaces act on their eigenkets as usual. Here we distribute over both sides of an equation, using `*` on the left-hand side, to see the product uncomputed, and `.` on the right-hand side to see it computed:

Release the product

The same operation but now using the dot product `.` operator. Start by delaying the operation

Recalling that this product is mathematically the same as (2.11), and that

by releasing the delayed product (2.12) we have

Reset hideketlabel

Entangled States and the Bell basis

With the introduction of disjointed Hilbert spaces in Maple it is possible to represent entangled quantum states in a simple way, basically as done with paper and pencil.

Recalling the Hilbert spaces set at this point are,

where  acts on the tensor product of the spaces where  and  act. A state of  can then always be written as

where  is a matrix of complex coefficients. Bra  states of  are formed as usual taking the Dagger

Example: the Bell basis for a system of two qubits

Consider a 2-dimensional space of states acted upon by the operator , and let  act upon another, disjointed, Hilbert space that is a replica of the Hilbert space on which  acts. Set the dimensions of ,  and  respectively equal to 2, 2 and 2x2 (see Setup)

The system C with the two subsystems A and B represents the a two qubits system. The standard basis for C can be constructed in a natural way from the basis of  Kets of A and B, , by taking their tensor products:

Set a more mathematical display for the imaginary unit

The four entangled Bell states also form a basis of C and are given by

There is no standard notation for denoting a Bell state (the linar combinations of the right-hand sides above). The convention used here relates to the definition of the Bell states related to the Pauli matrices shown below. Regardless fo the convention used, this basis is orthonormal. That can be verified by taking dot products, for example:

In steps, perform the same operation but using the star (`*`) operator, so that the contraction is represented but not performed

Evaluate now the result at `*` = `.`, that is transforming the star product into a dot product

The Bell basis and its relation with the Pauli matrices

The Bell basis can be constructed departing from  using the Pauli matrices . For that purpose, using a Vector representation for ,

Multiplying by each of the  Pauli Matrices and performing the matrix operations we have

In this result we see that  and  flip the state, transforming  into ,  also multiplies the state by the imaginary unit , while  leaves the state  unchanged.

We can rewrite all that by removeing from (3.19) the Vector representations of (3.17). For that purpose, create a list of substitution equations, replacing the Vectors by the Kets

So the action of  in  is given by

For , the same operations result in

To obtain the other three Bell states using the results (3.21) and (3.24), indicate to the system that the Pauli matrices operate in the subspace where  operates

Multiplying  given in (3.7) by each of the three  we get the other three Bell states

Substitute in this result the first equations of (3.21) and (3.24)

This is  defined in (3.8)

Multiplying now by  and substituting  using the  equations of (3.21) and (3.24) we get

The above is  defined in (3.9)

Finally, multiplying  by

Substituting

We get

which is

Entangled States, Operators and Projectors

Consider a fourth operator, , that is Hermitian and acts on the same space of , and then it has the same dimension,

To operate in a practical way with these operators, Bras and Kets, however, bracket rules reflecting their relationship are necessary. From the definition of  as acting on the tensor product of  spaces where  and  act (see (3.2)) and taking into account the dimensions specified for ,  and  we have

The bracket rules for ,  and  are the first two of these; Set these rules, so that the system can take them into account

If we now recompute (4.5), the left-hand side is also computed

Suppose now that you want to compute with the Hermitian operator , that operates on the same space as , both using C using the operators  and , as in

where  =  when  is a product (not entagled) state.

For  it suffices to set a bracket rule

After that, the system operates taking the rule into account

Regarding , since  belongs to the tensor product of spaces A and B, it can be an entangled operator, one that you cannot represent just as a product of one operator acting on A times another one acting on B. A computational representation for the operator  (that is not just itself or as abstract) is not possible in the general case. For that you can use a different feature: define the action of the operator  on Kets of  and .

Basically, we want:

A program sketch for that would be:

if H is applied to a Ket of A or B then

    if H itself is indexed then
        return H accumulating its indices, followed by the index of the Ket
    else

        return H indexed by the index of the Ket;
otherwise
    return the dot product operation uncomputed, unevaluated

In Maple language, that program-sketch becomes

Let's see it in action. Start erasing the Physics performance remember tables, that remember results like  computed before the definition of

Recalling that  is Hermitian,

Note that the definition of  as a procedure does not interfer with the setting of an bracket rule for it with , that is still working

but the definition takes precedence, so if in it you indicate what to do with a  Ket, that will be taken into account before the bracket rule. Finally, In the typical case, the first four results, (4.11), (4.12), (4.13) and (4.14) are operators while (4.15) is a scalar; you can always represent the scalar aspect by substituing the noncommutative operator  by a related scalar, say H.

Since the algebra rules for computing with eigenkets of ,  and  were already set in (4.6), from the projectors above you can construct any subspace projector, for example

The conjugate of  is due to the contraction or attachment from the right of (4.18), that is with

The coefficients  satisfy constraints due to the normalization of  Kets of  and . One can derive these contraints by inserting the unit operator  constructing this identity

Transform this result into a function P  to explore the identity further

The first and third indices refer to the quantum numbers of , the second and fourth to , so the the right-hand sides in the following are respectively 1 and 0

To get the whole system of equations satisfied by the coefficients , use P to construct an Array with four indices running from 0..1

Convert the whole Array into a set of equations

Reference

1. PDE and BC problems solved using linear change of variables

PDE and BC problems often require that the boundary and initial conditions be given at certain evaluation points (usually in which one of the variables is equal to zero). Using linear changes of variables, however, it is possible to change the evaluation points of BC, obtaining the solution for the new variables, and then changing back to the original variables. This is now automatically done by the pdsolve command.

Example 1: A heat PDE & BC problem in a semi-infinite domain:

Note the evaluation points A and B. The method typically described in textbooks requires the evaluation points to be . The change of variables automatically used in this case is:

so that pdsolve's task becomes solving this other problem, now with the appropriate evaluation points

and then changing the variables back to the original {x, t, u} and giving the solution. The process all in one go:

Example 2: A heat PDE with a source and a piecewise initial condition

Example 3: A wave PDE & BC problem in a semi-infinite domain:

Example 4: A wave PDE & BC problem in a semi-infinite domain:

Example 5: A wave PDE with a source:

2. It is now possible to specify or exclude method(s) for solving

The pdsolve/BC solving methods can now be indicated, either to be used for solving, as in  to be tried in the indicated order, or to be excluded, as in . The methods and sub-methods available are organized in a table,

So, for example, the methods for PDEs of first order and second order are, respectively,

Some methods have sub-methods (their existence is visible in (9)):

Example 6:

Example 7:

3. Series solutions for linear PDE and BC problems solved via product separation with eigenvalues that are the roots of algebraic expressions which cannot be inverted

Linear problems for which the PDE can be separated by product, giving rise to Sturm-Liouville problems for the separation constant (eigenvalue) and separated functions (eigenfunctions), do not always result in solvable equations for the eigenvalues. Below are examples where the eigenvalues are respectively roots of a sum of   functions and of the non-inversible equation .

Example 8: This problem represents the temperature distribution in a thin circular plate whose lateral surfaces are insulated (Articolo example 6.9.2):

In the above we see that the eigenvalue  satisfies . When  is the root of one single BesselJ or BesselY function of integer order, the Maple functions BesselJZeros and BesselYZeros are used instead. That is the case, for instance, if we slightly modify this problem changing the first boundary condition to be  instead of

Example 9: This problem represents the temperature distribution in a thin rod whose left end is held at a fixed temperature of 5 and whose right end loses heat by convection into a medium whose temperature is 10. There is also an internal heat source term in the PDE (Articolo's textbook, example 8.4.3):

For information on how to test or plot a solution like the one above, please see the end of the Mapleprimes post "Sturm-Liouville problem with eigenvalues that are the roots of algebraic expressions which cannot be inverted" 

4. Superposition method for linear PDE with more than one non-homogeneous BC

Previously, for linear homogeneous PDE problems with non-periodic initial and boundary conditions, pdsolve was only consistently able to solve the problem as long as at most one of those conditions was non-homogeneous. The superposition method works by taking advantage of the linearity of the problem and the fact that the solution to such a problem in which two or more of the BC are non-homogeneous can be given as

u =  + ...,  where each  is a solution of the PDE with all but one of the BC homogenized.

Example 10: A Laplace PDE with one homogeneous and three non-homogeneous conditions:

5. Polynomial solutions method:

This new method gives pdsolve better performance when the PDE & BC problems admit polynomial solutions.

Example 11:

6. Solving more problems using the Laplace transform or the Fourier transform

These methods now solve more problems and are no longer restricted to PDE of first or second order.

Example 12: A third order PDE & BC problem:

Example 13: A PDE & BC problem that is solved using Laplace transform:

To see the computational flow, the solving methods used and in which order they are tried use

Example 14:

Reset the infolevel to avoid displaying the computational flow:

7. Improvements to solving heat and wave PDE, with or without a source:

Example 15: A heat PDE:

To verify an infinite series solution such as this one you can first use pdetest

To verify that the last condition, for  is satisfied, we plot the first 1000 terms of the series solution with  and make sure that it coincides with the plot of  the right-hand side of the initial condition . Expected: the two plots superimpose each other

Example 16: A heat PDE in a semi-bounded domain:

Example 17: A wave PDE in a semi-bounded domain:

Example 18: A wave PDE with a source

Example 19: Another wave PDE with a source

8. Improvements in series methods for Laplace PDE problems

Example 21: A Laplace PDE for which we seek a solution that remains bounded as y approaches :