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These are Posts and Questions associated with the product, Maple

How do I plot a piece of the implicit curve y^4=y^2-x^2 (when x = 0.1 .. 0.3 and y = 0.1 .. 0.3) and the point (0.2, 0.2043096437) on the curve with a different color compared to the implicit curve? I want that both x axis and y axis represents the interval (-0.5,0.5).

I'm using Maple 18

 

Hi all,

I have some difficulties to get simple real expressions for those simple integrals with assumptions.

Download integrals.mw

NB: With mathematica, I get simple real expressions results.

Thanks

with ref to the following post
https://www.mapleprimes.com/questions/210280-Exporting--Graph-

I am trying to export a plot which contains a subscript in its axis label. The label displays correctly in the maple worksheet but when I export it to a jpg file the subscript gets replaced with square brackets. Kindly guide me how can I get rid of this problem

mprime.mw

Tensor product of Quantum States using Dirac's Bra-Ket Notation - 2018

 

There has been increasing interest in the details of the Maple implementation of tensor products using Dirac's notation, developed during 2018. Tensor products of Hilbert spaces and related quantum states are relevant in a myriad of situations in quantum mechanics, and in particular regarding quantum information. Below is a presentation up-to-date of the design and implementation, with input/output and examples, organized in four sections:

 

• 

The basic ideas and design implemented

• 

Tensor product notation and the hideketlabel option

• 

Entangled States and the Bell basis

• 

Entangled States, Operators and Projectors

 

Part of this development is present in Maple 2018.2. To reproduce what you see below, however, you need a more recent version, as the one distributed within the Maplesoft Physics Updates (version 272 or higher).

 

The basic ideas and design implemented

 

 

Suppose A and B are quantum operators and Ket(A, n), et(B, m) are, respectively, their eigenkets. The following works since the introduction of the Physics package in Maple

with(Physics)

Setup(op = {A, B})

`* Partial match of  '`*op*`' against keyword '`*quantumoperators*`' `

 

_______________________________________________________

 

[quantumoperators = {A, B}]

(1.1)

A*Ket(A, alpha) = A.Ket(A, alpha)

Physics:-`*`(A, Physics:-Ket(A, alpha)) = alpha*Physics:-Ket(A, alpha)

(1.2)

B*Ket(B, beta) = B.Ket(B, beta)

Physics:-`*`(B, Physics:-Ket(B, beta)) = beta*Physics:-Ket(B, beta)

(1.3)

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

 

1) To represent that situation, a new keyword in Setup , hilbertspaces, is introduced. With it you can indicate the quantum operators that act on a Hilbert space, say as in hilbertdspaces = {{A}, {B}} with the meaning that the operator A acts on one Hilbert space while B 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 - hilbertdspaces = {{A}, {B}}, automatically, A, B become quantum operators satisfying (see comment (ii) on page 156 of ref.[1])

 

"[A,B][-]=0"

 

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]):

 

`⊗`(Ket(A, alpha), Ket(B, beta)) = `⊗`(Ket(B, beta), Ket(A, alpha)) 

 

`⊗`(Bra(A, alpha), Ket(B, beta)) = `⊗`(Ket(B, beta), Bra(A, alpha)) 

 

while

Bra(A, alpha)*Ket(A, alpha) <> Bra(A, alpha)*Ket(A, alpha)

 

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

 

A*Ket(B, n) = A*Ket(B, n)

  

and the same for the Dagger of this equation, that is

Bra(B, n)*Dagger(A) = Bra(B, n)*Dagger(A)

 

  

Hence, when we write the left-hand sides of the two equations above and press enter, they are automatically rewritten (returned) as the right-hand sides.

 

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 products formed with operators, or Bras and Kets belonging to different Hilbert spaces (set as such using Setup  and the keyword hilbertspaces), are now displayed with the symbol 5 in between, as in Ket(A, n)*Ket(B, n) instead of Ket(A, n)*Ket(B, n), and `&otimes;`(A, B) instead of A*B. The product of an operator A of one space and a KetNULL of another space Ket(B, n) however, is displayed AA, without 5.

• 

A new Setup option hideketlabel , makes all the labels in Kets and Bras to be hidden at the time of displaying Kets, Bras and Bracket, so when you set it entering Setup(hideketlabel = true),

 "Ket(A,m,n,l"  

  

is displayed as

Ket(A, m, n, l)

 

  

This is the notation frequently used when working with angular momentum or in quantum information, where tensor products of Hilbert spaces are used.

Design details

   

Tensor product notation and the hideketlabel option

 

 

According to the design section, set now two disjointed Hilbert spaces with operators A, C acting on one of them and B, C on the other one (you can think of  C = `&otimes;`(A, B))

 

Setup(hilbertspaces = {{A, C}, {B, C}})

[disjointedspaces = {{A, C}, {B, C}}]

(2.1)

 

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

Ket(A, 1)*Ket(B, 0)

Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0))

(2.2)
• 

As explained in the Details of the design section, the ordering of the Hilbert spaces in tensor products is now preserved: Bras (Kets) of the first space always appear before Bras (Kets) of the second space. For example, construct a projector into the state (2.2)

Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0))*Dagger(Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0)))

Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

(2.3)

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, `&otimes;`(Ket(A, 1)*Bra(A, 1), `&otimes;`(Ket(B, 0), Bra(B, 0))) instead of the projector sorting style of  (2.3). Both reorderings of Kets and Bras are mathematically equal.

 

• 

Because that ordering is preserved, one can now hide the label of Bras and Kets without ambiguity, as it is usual in textbooks (e.g. in Quantum Information). For that purpose use the new keyword option hideketlabel

Setup(hide = true)

`* Partial match of  '`*hide*`' against keyword '`*hideketlabel*`' `

 

_______________________________________________________

 

[hideketlabel = true]

(2.4)

The display for (2.3) is now

Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))

Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

(2.5)

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)

show

Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

(2.6)

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:

(`*` = `.`)(A, Ket(A, 1))

Physics:-`*`(A, Physics:-Ket(A, 1)) = Physics:-Ket(A, 1)

(2.7)

(`*` = `.`)(A, Ket(A, 0))

Physics:-`*`(A, Physics:-Ket(A, 0)) = 0

(2.8)
• 

The tensor product of operators belonging to different Hilbert spaces is also displayed using 5

A*B

Physics:-`*`(A, B)

(2.9)
• 

 As mentioned in the preceding design section, using the commutativity between operators, Bras and Kets that belong to different Hilbert spaces, within a product, operators are placed contiguous to the Kets and Bras belonging to the space where the operator acts. For example, consider the delayed product represented using the start `*` operator

'Physics[`*`](A, B)*Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))'

Physics:-`*`(A, B, Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

(2.10)

Release the product

%

Physics:-`*`(A, Physics:-Ket(A, 1), B, Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

(2.11)

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

'Physics[`*`](A, B).Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))'

Parse:-ConvertTo1D, "invalid input %1", Typesetting:-mprintslash([A*B.Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))], [A*B.Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))])

(2.12)

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

B.Ket(B, 0)

0

(2.13)

by releasing the delayed product (2.12) we have

Typesetting[delayDotProduct](Physics[`*`](A, B), Physics[`*`](Ket(A, 1), Ket(B, 0), Bra(A, 1), Bra(B, 0)))

0

(2.14)

Reset hideketlabel

Setup(hideketlabel = false)

[hideketlabel = false]

(2.15)

Implementation details

   

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,

Setup(hilbert)

`* Partial match of  '`*hilbert*`' against keyword '`*hilbertspaces*`' `

 

_______________________________________________________

 

[disjointedspaces = {{A, C}, {B, C}}]

(3.1)

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

Ket(C, m, n) = Sum(Sum(M[j, p]*Ket(A, j)*Ket(B, p), j), p)

Physics:-Ket(C, m, n) = Sum(Sum(M[j, p]*Physics:-`*`(Physics:-Ket(A, j), Physics:-Ket(B, p)), j), p)

(3.2)

where M[j, p] is a matrix of complex coefficients. Bra  states of C are formed as usual taking the Dagger

Dagger(Ket(C, m, n) = Sum(Sum(M[j, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j), p))

Physics:-Bra(C, m, n) = Sum(Sum(conjugate(M[j, p])*Physics:-`*`(Physics:-Bra(A, j), Physics:-Bra(B, p)), j), p)

(3.3)

 

• 

By definition, all states Ket(C, alpha, beta) that can be written exactly as `&otimes;`(Ket(A, alpha), Ket(B, beta)), that is, the product of a arbitrary state of the subspace A and another of the subspace B, are product states, and all the other ones are entangled states. Entangelment is a property that is independent of the basis `&otimes;`(Ket(A, j), Ket(B, p))used in (3.2).

The physical interpretation is the standard one: when the state of a system constituted by two subsystems A and B is represented by a product state, the properties of the subsystem A are well defined and all given by "Ket(A,alpha),"while those for the subsystem B by NULL. When the system is in an entangled state one typically cannot assign definite properties to the individual subsystems A or B, each subsystem has no independent reality.

To determine whether a state Ket(C, alpha, beta) is or not entangled it then suffices to check the rank R of the matrix M[j, p] (see LinearAlgebra:-Rank ): when R = 1 the state is a product state, otherwise it is an entangled state. When the state being analized belongs to the tensor product of two subspaces, R = 1.is equivalent to having the determinant of M[j, p] equal to 0. The condition R = 1, however, is more general, and suffices to determine whether a state is a product state also on a Hilbert space that is the tensor product of three or more subspaces: "`&Hscr;`^()=`&Hscr;`^((1))&otimes;`&Hscr;`^((2))&otimes;`&Hscr;`^((3))... `&Hscr;`^((n))", in which case the matrix M will have more rows and columns and a determinant equal to 0 would only warrant the possibility of factorizing one Ket.

 

Example: the Bell basis for a system of two qubits

 

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

Setup(quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2})

[quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2}]

(3.4)

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, {Ket(A, 0), Ket(A, 1), Ket(B, 0), Ket(B, 1)}, by taking their tensor products:

seq(seq(Ket(A, j)*Ket(B, k), k = 0 .. 1), j = 0 .. 1)

Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1)), Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)), Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1))

(3.5)

Set a more mathematical display for the imaginary unit

interface(imaginaryunit = i)

 

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

Setup(op = `&Bscr;`)

`* Partial match of  '`*op*`' against keyword '`*quantumoperators*`' `

 

_______________________________________________________

 

[quantumoperators = {`&Bscr;`, A, B, C, E}]

(3.6)

Ket(`&Bscr;`, 0) = (Ket(A, 0)*Ket(B, 0)+Ket(A, 1)*Ket(B, 1))/('sqrt')(2)

Physics:-Ket(`&Bscr;`, 0) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))/sqrt(2)

(3.7)

Ket(`&Bscr;`, 1) = (Ket(A, 0)*Ket(B, 1)+Ket(A, 1)*Ket(B, 0))/('sqrt')(2)

Physics:-Ket(`&Bscr;`, 1) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))/sqrt(2)

(3.8)

Ket(`&Bscr;`, 2) = i*(Ket(A, 0)*Ket(B, 1)-Ket(A, 1)*Ket(B, 0))/('sqrt')(2)

Physics:-Ket(`&Bscr;`, 2) = I*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))/sqrt(2)

(3.9)

Ket(`&Bscr;`, 3) = (Ket(A, 0)*Ket(B, 0)-Ket(A, 1)*Ket(B, 1))/('sqrt')(2)

Physics:-Ket(`&Bscr;`, 3) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))/sqrt(2)

(3.10)

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:

Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)).(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

1 = 1

(3.11)

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

Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

Physics:-`*`(Physics:-Bra(`&Bscr;`, 0), Physics:-Ket(`&Bscr;`, 0)) = (1/2)*Physics:-`*`(Physics:-`*`(Physics:-Bra(A, 0), Physics:-Bra(B, 0))+Physics:-`*`(Physics:-Bra(A, 1), Physics:-Bra(B, 1)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.12)

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

eval(Physics[`*`](Bra(`&Bscr;`, 0), Ket(`&Bscr;`, 0)) = (1/2)*Physics[`*`](Physics[`*`](Bra(A, 0), Bra(B, 0))+Physics[`*`](Bra(A, 1), Bra(B, 1)), Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))), `*` = `.`)

1 = 1

(3.13)

Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))*(Ket(`&Bscr;`, 1) = (Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2))

Physics:-`*`(Physics:-Bra(`&Bscr;`, 0), Physics:-Ket(`&Bscr;`, 1)) = (1/2)*Physics:-`*`(Physics:-`*`(Physics:-Bra(A, 0), Physics:-Bra(B, 0))+Physics:-`*`(Physics:-Bra(A, 1), Physics:-Bra(B, 1)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

(3.14)

eval(Physics[`*`](Bra(`&Bscr;`, 0), Ket(`&Bscr;`, 1)) = (1/2)*Physics[`*`](Physics[`*`](Bra(A, 0), Bra(B, 0))+Physics[`*`](Bra(A, 1), Bra(B, 1)), Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))), `*` = `.`)

0 = 0

(3.15)

The Bell basis and its relation with the Pauli matrices

 

The Bell basis can be constructed departing from Ket(`&Bscr;`, 0) using the Pauli matrices sigma[j]. For that purpose, using a Vector representation for Ket(A, j),

Physics:-Ket(`&Bscr;`, 0)

(3.16)

Ket(B, 0) = Vector([1, 0]), Ket(B, 1) = Vector([0, 1])

Physics:-Ket(B, 0) = Vector[column](%id = 18446744078301209294), Physics:-Ket(B, 1) = Vector[column](%id = 18446744078301209414)

(3.17)

Multiplying Ket(B, 0)by each of the sigma[j] Pauli Matrices and performing the matrix operations we have

"[seq(Psigma[j] . ?[1], j=1..3)]"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Psigma[1].Vector[column](%id = 18446744078301209294), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = Physics:-Psigma[2].Vector[column](%id = 18446744078301209294), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Psigma[3].Vector[column](%id = 18446744078301209294)]

(3.18)

"map(u -> lhs(u) =Library:-PerformMatrixOperations(rhs(u)),?)"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376366918), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376368838), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376358606)]

(3.19)

In this result we see that sigma[1] and sigma[2] flip the state, transforming Ket(B, 0) into Ket(B, 1), sigma[2] also multiplies the state by the imaginary unit I, while sigma[3] leaves the state Ket(B, 0) 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

"map(rhs = lhs,[?, i *~ ?])"

[Vector[column](%id = 18446744078301209294) = Physics:-Ket(B, 0), Vector[column](%id = 18446744078301209414) = Physics:-Ket(B, 1), Vector[column](%id = 18446744078376351494) = I*Physics:-Ket(B, 0), Vector[column](%id = 18446744078376351734) = I*Physics:-Ket(B, 1)]

(3.20)

So the action of sigma[j] in Ket(B, 0) is given by

"Library:-SubstituteMatrix(?,?)"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = I*Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Ket(B, 0)]

(3.21)

For Ket(B, 1), the same operations result in

"[seq(Psigma[j] . ?[2], j=1..3)]"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Psigma[1].Vector[column](%id = 18446744078301209414), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = Physics:-Psigma[2].Vector[column](%id = 18446744078301209414), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = Physics:-Psigma[3].Vector[column](%id = 18446744078301209414)]

(3.22)

"map(u -> lhs(u) =Library:-PerformMatrixOperations(rhs(u)),?)"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464860518), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464862438), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464856182)]

(3.23)

"Library:-SubstituteMatrix(?,?)"

[Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = -I*Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = -Physics:-Ket(B, 1)]

(3.24)

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 B operates

Setup(hilbert = {{B, C, Psigma}})

`* Partial match of  '`*hilbert*`' against keyword '`*hilbertspaces*`' `

 

_______________________________________________________

 

[disjointedspaces = {{A, C}, {B, C, Physics:-Psigma}}]

(3.25)

 

Multiplying Ket(`&Bscr;`, 0) given in (3.7) by each of the three sigma[j] we get the other three Bell states

Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)

Physics:-Ket(`&Bscr;`, 0) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.26)

Psigma[1]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[1], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.27)

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

[Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][1], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][1]

Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Ket(B, 0)

(3.28)

map(rhs = lhs, [Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0)])

[Physics:-Ket(B, 1) = Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)), Physics:-Ket(B, 0) = Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1))]

(3.29)

subs([Ket(B, 1) = Physics[`*`](Physics[Psigma][1], Ket(B, 0)), Ket(B, 0) = Physics[`*`](Physics[Psigma][1], Ket(B, 1))], Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][1], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))

Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[1], Physics:-`*`(Physics:-Ket(A, 0), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0))))

(3.30)

factor(Simplify(Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][1], Physics[`*`](Ket(A, 0), Physics[`*`](Physics[Psigma][1], Ket(B, 1)))+Physics[`*`](Ket(A, 1), Physics[`*`](Physics[Psigma][1], Ket(B, 0))))))

Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

(3.31)

This is Ket(`&Bscr;`, 1) defined in (3.8)

Ket(`&Bscr;`, 1) = (Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2)

Physics:-Ket(`&Bscr;`, 1) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

(3.32)

(Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))))-(Ket(`&Bscr;`, 1) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))))

Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 1) = 0

(3.33)

Multiplying now by sigma[2] and substituting Ket(B, j) using the 2^nd equations of (3.21) and (3.24) we get Ket(`&Bscr;`, 1)

Psigma[2]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[2], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.34)

[Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][2], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][2]

Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = I*Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = -I*Physics:-Ket(B, 0)

(3.35)

zip(isolate, [Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0)], [Ket(B, 1), Ket(B, 0)])

[Physics:-Ket(B, 1) = -I*Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)), Physics:-Ket(B, 0) = I*Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1))]

(3.36)

factor(Simplify(subs([Ket(B, 1) = -I*Physics[`*`](Physics[Psigma][2], Ket(B, 0)), Ket(B, 0) = I*Physics[`*`](Physics[Psigma][2], Ket(B, 1))], Physics[`*`](Physics[Psigma][2], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][2], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))))

Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0)) = ((1/2)*I)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

(3.37)

The above is Ket(`&Bscr;`, 2) defined in (3.9)

Ket(`&Bscr;`, 2) = I*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2)

Physics:-Ket(`&Bscr;`, 2) = ((1/2)*I)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

(3.38)

Expand((Physics[`*`](Physics[Psigma][2], Ket(`&Bscr;`, 0)) = ((1/2)*I)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0))))-(Ket(`&Bscr;`, 2) = ((1/2)*I)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0)))))

Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 2) = 0

(3.39)

Finally, multiplying Ket(`&Bscr;`, 2) by sigma[3]

Psigma[3]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[3], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.40)

Substituting

[Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][3], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][3]

Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = -Physics:-Ket(B, 1)

(3.41)

(rhs = lhs)((Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1))[1]), (rhs = lhs)(-(Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1))[2])

Physics:-Ket(B, 0) = Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)), Physics:-Ket(B, 1) = -Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1))

(3.42)

We get ``

factor(Simplify(subs(Ket(B, 0) = Physics[`*`](Physics[Psigma][3], Ket(B, 0)), Ket(B, 1) = -Physics[`*`](Physics[Psigma][3], Ket(B, 1)), Physics[`*`](Physics[Psigma][3], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][3], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))))

Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.43)

which is Ket(`&Bscr;`, 2)

Ket(`&Bscr;`, 3) = (Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)

Physics:-Ket(`&Bscr;`, 3) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

(3.44)

Expand((Physics[`*`](Physics[Psigma][3], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1))))-(Ket(`&Bscr;`, 3) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1)))))

Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 3) = 0

(3.45)

Entangled States, Operators and Projectors

 

 

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

Setup(additionally, hermitian = H, basisdimension = {H[1] = 2, H[2] = 2}, hilbertspaces = {{A, C, H}, {B, C, H}})

`* Partial match of  '`*hermitian*`' against keyword '`*hermitianoperators*`' `

 

`* Partial match of  '`*basisdimension*`' against keyword '`*quantumbasisdimension*`' `

 

_______________________________________________________

 

[disjointedspaces = {{A, C, H}, {B, C, H}, {B, C, Physics:-Psigma}}, hermitianoperators = {H}, quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2, H[1] = 2, H[2] = 2}]

(4.1)

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

Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Ket(A, j)*Ket(B, p), j = 0 .. 1), p = 0 .. 1)

Physics:-Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics:-`*`(Physics:-Ket(A, j), Physics:-Ket(B, p)), j = 0 .. 1), p = 0 .. 1)

(4.2)

Bra(A, k).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

Physics:-Bracket(Physics:-Bra(A, k), Physics:-Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1)

(4.3)

Bra(B, k).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

Physics:-Bracket(Physics:-Bra(B, k), Physics:-Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1)

(4.4)

Bra(A, k).Bra(B, l).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

Physics:-`*`(Physics:-Bra(A, k), Physics:-Bracket(Physics:-Bra(B, l), Physics:-Ket(C, a, b))) = M[a, k, b, l]

(4.5)

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

Setup(Bracket(Bra(A, k), Ket(C, a, b)) = Sum(M[a, k, b, p]*Ket(B, p), p = 0 .. 1), Bracket(Bra(B, k), Ket(C, a, b)) = Sum(M[a, j, b, k]*Ket(A, j), j = 0 .. 1))

[bracketrules = {%Bracket(%Bra(A, k), %Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1), %Bracket(%Bra(B, k), %Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1)}]

(4.6)

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

Bracket(C, i, j, H, C, k, l) = `&Hscr;`

Physics:-Bracket(Physics:-Bra(C, I, j), H, Physics:-Ket(C, k, l)) = `&Hscr;`

(4.7)

Bra(A, k).Bra(B, l).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

M[a, k, b, l] = M[a, k, b, l]

(4.8)

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

 

Bracket(Bra(C, I, j), H, Ket(C, k, l)) = `&Hscr;`[i, j, k, l]

 

`&otimes;`(Bra(A, I), Bra(B, j))*H*`&otimes;`(Ket(A, k), Ket(B, l)) = H[I, j, k, l]

 

where `&Hscr;`[i, j, k, l] = H[I, j, k, l] when Ket(C, a, b) is a product (not entagled) state.

 

For Bracket(Bra(C, I, j), H, Ket(C, k, l)) = `&Hscr;`[I, j, k, l] it suffices to set a bracket rule

Setup(%Bracket(Bra(C, a, b), H, Ket(C, c, d)) = `&Hscr;`[a, b, c, d], real = `&Hscr;`)

`* Partial match of  '`*real*`' against keyword '`*realobjects*`' `

 

_______________________________________________________

 

[bracketrules = {%Bracket(%Bra(A, k), %Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1), %Bracket(%Bra(B, k), %Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1), %Bracket(%Bra(C, a, b), H, %Ket(C, c, d)) = `&Hscr;`[a, b, c, d]}, realobjects = {`&Hscr;`, x, y, z}]

(4.9)

After that, the system operates taking the rule into account

Bra(C, j, k).H.Ket(C, m, n)

`&Hscr;`[j, k, m, n]

(4.10)

Regarding `&otimes;`(Bra(A, I), Bra(B, j))*H*`&otimes;`(Ket(A, k), Ket(B, l)) = H[I, j, k, l]NULL, since H 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 Bra(B, j)*H*Ket(A, k) (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 H on Kets of A and B.

 

Basically, we want:

 

"H*Ket(A,k)-> H[k]"

"H[k]*Ket(B,l)->H[k,l]"

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

 

"H := K ->   if K::Ket and op(1, K)::'identical(A,B)' then      if procname::'indexed' then         if nops(procname) <4 then             H[op(procname), op(2, K)]    #` accumulate indices`         else             'H . K'         fi     else          H[op(2, K)]     fi  else      'procname . K'  fi:"

 

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

 

Library:-Forget()

H.Ket(A, k)

H[k]

(4.11)

Recalling that H is Hermitian,

Bra(B, j).H

H[j]

(4.12)

Bra(B, j).H.Ket(A, k)

H[j, k]

(4.13)

Bra(B, j).H.Ket(A, k).Ket(B, l)

H[j, k, l]

(4.14)

Bra(A, i).Bra(B, j).H.Ket(A, k).Ket(B, l)

H[I, j, k, l]

(4.15)

Note that the definition of H as a procedure does not interfer with the setting of an bracket rule for it with Ket(C, a, b), that is still working

Bra(C, i, j).H.Ket(C, k, l)

`&Hscr;`[I, j, k, l]

(4.16)

but the definition takes precedence, so if in it you indicate what to do with a C 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 H by a related scalar, say H.

 

• 

You can set the projectors for all these operators / spaces. For example,

`&Iopf;__A` := Projector(Ket(A, i)); `&Iopf;__B` := Projector(Ket(B, i)); `&Iopf;__C` := Projector(Ket(C, a, b))

Sum(Physics:-`*`(Physics:-Ket(A, n), Physics:-Bra(A, n)), n = 0 .. 1)

 

Sum(Physics:-`*`(Physics:-Ket(B, n), Physics:-Bra(B, n)), n = 0 .. 1)

 

Sum(Sum(Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(C, a, b)), a = 0 .. 1), b = 0 .. 1)

(4.17)

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

Bra(A, m).`&Iopf;__C`

Sum(Sum(Sum(M[a, m, b, p]*Physics:-`*`(Physics:-Ket(B, p), Physics:-Bra(C, a, b)), p = 0 .. 1), a = 0 .. 1), b = 0 .. 1)

(4.18)

`&Iopf;__C`.Ket(A, m)

Sum(Sum(Sum(conjugate(M[a, m, b, p])*Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(B, p)), p = 0 .. 1), a = 0 .. 1), b = 0 .. 1)

(4.19)

The conjugate of M[a, m, b, p] is due to the contraction or attachment from the right of (4.18), that is with

Dagger(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

Physics:-Bra(C, a, b) = Sum(Sum(conjugate(M[a, j, b, p])*Physics:-`*`(Physics:-Bra(A, j), Physics:-Bra(B, p)), j = 0 .. 1), p = 0 .. 1)

(4.20)

 

The coefficients M[a, m, b, p] satisfy constraints due to the normalization of  Kets of A and B. One can derive these contraints by inserting the unit operator `#msub(mi("&Iopf;"),mi("C"))` constructing this identity

Sum(Sum(Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(C, a, b)), a = 0 .. 1), b = 0 .. 1)

(4.21)

Bra(A, m).Bra(B, n).`&Iopf;__C`.Ket(A, r).Ket(B, s) = Bra(A, m).Bra(B, n).Ket(A, r).Ket(B, s)

Sum(Sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics:-KroneckerDelta[m, r]*Physics:-KroneckerDelta[n, s]

(4.22)

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

P := unapply(subs(Sum = sum, Sum(Sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics[KroneckerDelta][m, r]*Physics[KroneckerDelta][n, s]), m, n, r, s)

proc (m, n, r, s) options operator, arrow; sum(sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics:-KroneckerDelta[m, r]*Physics:-KroneckerDelta[n, s] end proc

(4.23)

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

P(1, 0, 1, 0)

conjugate(M[0, 1, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 1, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 1, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 1, 1, 0])*M[1, 1, 1, 0] = 1

(4.24)

P(1, 0, 0, 0)

conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 0] = 0

(4.25)

To get the whole system of equations satisfied by the coefficients M[a, m, b, n], use P to construct an Array with four indices running from 0..1

Array(`$`(0 .. 1, 4), P)

_rtable[18446744078376377150]

(4.26)

Convert the whole Array into a set of equations

"simplify(convert(Typesetting:-msub(Typesetting:-mi("_rtable",italic = "true",mathvariant = "italic"),Typesetting:-mrow(Typesetting:-mn("18446744078376377150",mathvariant = "normal")),subscriptshift = "0"),setofequations))"

{abs(M[0, 0, 0, 0])^2+abs(M[1, 0, 0, 0])^2+abs(M[0, 0, 1, 0])^2+abs(M[1, 0, 1, 0])^2 = 1, abs(M[0, 0, 0, 1])^2+abs(M[1, 0, 0, 1])^2+abs(M[0, 0, 1, 1])^2+abs(M[1, 0, 1, 1])^2 = 1, abs(M[0, 1, 0, 0])^2+abs(M[1, 1, 0, 0])^2+abs(M[0, 1, 1, 0])^2+abs(M[1, 1, 1, 0])^2 = 1, abs(M[0, 1, 0, 1])^2+abs(M[1, 1, 0, 1])^2+abs(M[0, 1, 1, 1])^2+abs(M[1, 1, 1, 1])^2 = 1, conjugate(M[0, 0, 0, 0])*M[0, 0, 0, 1]+conjugate(M[1, 0, 0, 0])*M[1, 0, 0, 1]+conjugate(M[0, 0, 1, 0])*M[0, 0, 1, 1]+conjugate(M[1, 0, 1, 0])*M[1, 0, 1, 1] = 0, conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 0] = 0, conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 1]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 1]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 1]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 1] = 0, conjugate(M[0, 0, 0, 1])*M[0, 0, 0, 0]+conjugate(M[1, 0, 0, 1])*M[1, 0, 0, 0]+conjugate(M[0, 0, 1, 1])*M[0, 0, 1, 0]+conjugate(M[1, 0, 1, 1])*M[1, 0, 1, 0] = 0, conjugate(M[0, 0, 0, 1])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 1])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 1])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 1])*M[1, 1, 1, 0] = 0, conjugate(M[0, 0, 0, 1])*M[0, 1, 0, 1]+conjugate(M[1, 0, 0, 1])*M[1, 1, 0, 1]+conjugate(M[0, 0, 1, 1])*M[0, 1, 1, 1]+conjugate(M[1, 0, 1, 1])*M[1, 1, 1, 1] = 0, conjugate(M[0, 1, 0, 0])*M[0, 0, 0, 0]+conjugate(M[1, 1, 0, 0])*M[1, 0, 0, 0]+conjugate(M[0, 1, 1, 0])*M[0, 0, 1, 0]+conjugate(M[1, 1, 1, 0])*M[1, 0, 1, 0] = 0, conjugate(M[0, 1, 0, 0])*M[0, 0, 0, 1]+conjugate(M[1, 1, 0, 0])*M[1, 0, 0, 1]+conjugate(M[0, 1, 1, 0])*M[0, 0, 1, 1]+conjugate(M[1, 1, 1, 0])*M[1, 0, 1, 1] = 0, conjugate(M[0, 1, 0, 0])*M[0, 1, 0, 1]+conjugate(M[1, 1, 0, 0])*M[1, 1, 0, 1]+conjugate(M[0, 1, 1, 0])*M[0, 1, 1, 1]+conjugate(M[1, 1, 1, 0])*M[1, 1, 1, 1] = 0, conjugate(M[0, 1, 0, 1])*M[0, 0, 0, 0]+conjugate(M[1, 1, 0, 1])*M[1, 0, 0, 0]+conjugate(M[0, 1, 1, 1])*M[0, 0, 1, 0]+conjugate(M[1, 1, 1, 1])*M[1, 0, 1, 0] = 0, conjugate(M[0, 1, 0, 1])*M[0, 0, 0, 1]+conjugate(M[1, 1, 0, 1])*M[1, 0, 0, 1]+conjugate(M[0, 1, 1, 1])*M[0, 0, 1, 1]+conjugate(M[1, 1, 1, 1])*M[1, 0, 1, 1] = 0, conjugate(M[0, 1, 0, 1])*M[0, 1, 0, 0]+conjugate(M[1, 1, 0, 1])*M[1, 1, 0, 0]+conjugate(M[0, 1, 1, 1])*M[0, 1, 1, 0]+conjugate(M[1, 1, 1, 1])*M[1, 1, 1, 0] = 0}

(4.27)

Reference

   

NULL


 

Download Tensor_Products_of_Quantum_States_-_2018.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi every body:

I have six ODEs in two case with their initial conditions: 

case 1:

eqs1:= diff(r[1](t), t, t)+.68041478*(diff(r[1](t), t))+732.61274*r[1](t)-0.11751824e-2*V(t) = -0.52084940e-2*q(t), 2.03*10^(-8)*(diff(V(t), t))+4.0650406*10^(-7)*V(t)+0.11751824e-2*(diff(r[1](t), t)) = 0, diff(q(t), t, t)+(9.5199780*(q(t)^2-1))*(diff(q(t), t))+1006.9998*q(t) = -2192.5091*(diff(r[1](t), t, t));
ics1 := V(0) = 0, q(0) = 0.1e-2, (D(q))(0) = 0, r[1](0) = 0, (D(r[1]))(0) = 0;

case 2:

eqs2:= diff(r[2](t), t, t)+3.3737665*(diff(r[2](t), t))+52672.540*r[2](t)-0.93177795e-2*V(t) = 0.18358493e-2*q(t), 2.03*10^(-8)*(diff(V(t), t))+4.0650406*10^(-7)*V(t)+0.11751824e-2*(diff(r[1](t), t))+0.93177795e-2*(diff(r[2](t), t)) = 0, diff(q(t), t, t)+(9.5199780*(q(t)^2-1))*(diff(q(t), t))+1006.9998*q(t) = -2192.5091*(diff(r[1](t), t, t))+772.79848*(diff(r[2](t), t, t));

ics2 := V(0) = 0, q(0) = 0.1e-2, (D(q))(0) = 0, r[2](0) = 0, (D(r[2]))(0) = 0;

How can i solve numeric stiff the ODEs and plot V(t) in case 2 for t=0..2?

tnx...

Hey there,

I've been struggling for a few days to integrate a function using rkf45.

For some context, I wish to numerically intergrate a function K * s_R (K,a,w) dK using runge kutta over a 2D grid of (a,w) values. Essentialy, performing the same sort of K-integral many different times for slightly different combinations of a and w. For one such (a,w) combination, the function looks something like this:

and rkf45 works just fine, and returns a numeric result. On the other hand, for a different combination of (a,w), the function looks like this:

and an attempt to perform rkf45 returns the following error: "cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up."

Here is a simplified version of my maple document for reference MaplePrimePleaseHelp.mw.

I'm happy to clarify anything that may be confusing and any thoughts or suggestions would be greatly appreciated.

Thank you!

please guide me how to write a simple code for Newton method to solve the following system of nonlinear equations
D= diagonal of A
L= lower triangular
U= upper Triangular
 


 

"n:=10;   x[0]:= ([1,2,3])    f[1]:=x[1]^(2)sin(x[1]x[2])+x[2]^(3)=5    f[2]:=e^(x[1]^(2))x[2]^(2)+3 x[1]^(2)x[2]+x[2]^(2)-6=0    newton method is     for k from 1 to n do    x[k+1]:=D^((-1))(L+U)*x[k]-D^((-1)b)    end do"

``


 

Download NM.mw
 

"n:=10;   x[0]:= ([1,2,3])    f[1]:=x[1]^(2)sin(x[1]x[2])+x[2]^(3)=5    f[2]:=e^(x[1]^(2))x[2]^(2)+3 x[1]^(2)x[2]+x[2]^(2)-6=0    newton method is     for k from 1 to n do    x[k+1]:=D^((-1))(L+U)*x[k]-D^((-1)b)    end do"

``


 

Download NM.mw

 

 

1) I want to visualize the probabilities of normal law. I used a technique ( a little heavy) ( see appendix) and I wonder if there is a way to do better ?

2 ) How do I calculate the probabilities of p( a<X<b) and p( X>c / X<d ) ?

3) How do I calculate the inverse probability in Maple. I only find the inverf(p) command under Maple TA?

NormaleVisualizationCourbe.mw

partial differential equation question??

utt = 9uxx

boundary conditions : u (0,t) = (2,t) = 0

initial conditions : u (x,0) = f(x) = { x(2-x)   ,0<x<1     ,      x   ,1<x<2  = 0 , t>0 ,

                         u(x,0) = 0 , 0<x<2

I have a physics question I need to program it by Maple , Can you help me to solve ? Its Problem 9.34 from Griffiths

Hi,

I'm a beginner in programing with Maple.

Maple has an algorithm "LinearSolve". 

I'm trying - just for fun - to generate an algorithm to solve a linear system of equations according to the Gauss-Jordan-method.

The linear System is given by:

M1*x=b

(in the worksheet: M1 is a matrix, point (1)

b is a vector, point (2) and x is a vector too and is the solution of system, point (3)).

Point (3) is to check - after the generated algorithm points out the solution - if the algorthm is working fine.

The augmented matrix M2 is given by <M1|b>, point (4).

First, I want to generate zeros under the main diagonal of matrix M1 in the augmented matrix M2.

In the worksheet there is my idea to do this, but I get a (perhaps simple) error message ("Error, Invalid if statement termination"). But I don't know what I'm doing wrong...

Perhaps someone can have a look on my idea (link to the file) and give me a hint?

Thanks a lot for your help!!

LGS_Gauss-Jordan.mw

I am working on an Isoperimetric problem.

I have two différential dL and dA. And I am seeking the function y for wich the two différentials are collinear.

So I want "dL/dA=constant"

 

I tried many expression, but I don't know how to express the "= constant"
I tried also (dL/dA)'=0 but no workable answer...

 

Thank's for your help !

Hi!

I have the following problem

> restart:with(plots): Digits := 60:
> A:=42403:
> B:=269.95:
> C:=269.95:
> J1:=0.55:
> J2:=1.02:
> a:=196200:
> b:=14342220:
> c:=1589220:
> k:=157.08/0.0875:
> sys := {A-B*diff(x(t),t)=J1*diff(x(t),t,t)+C*(x(t)-y(t)),k*(J2*diff(y(t),t,t)-C*(x(t)-y(t)))=-(a+b*(diff(y(t),t)^2)/k^2+c*diff(y(t),t,t)),  D(x)(0) = 157.08, D(y)(0)=0, D(y)(10)=157.08,  D(x)(10) = 154.98}:
> dsn1 := dsolve(sys,numeric):
Error, (in dsolve/numeric/BVPSolve) matrix is singular

I would like to extract the unique values of a few sets in a list (L1) and insert them in another list (L2) without the sets.

L1:=[{3,5},{5,4}]:

L2:=[3,5,4]:

Could anyone help me with a bit of code so this works for all values in sets in a list?

 

add(add(1,i=1..j),j=1..4);

10

(1)

S:=%add(%add(1,i=1..j),j=1..4);

%add(%add(1, i = 1 .. j), j = 1 .. 4)

(2)

value(S);

Error, (in value) unable to execute add

 

It seems that the inner add is evaluated prematurely.

 

eval(subs(%add=add, S));  # ok

10

(3)

subs[eval](%add=add, S);  # ?

Error, unable to execute add

 

 

 

A strange fact about filenames. 
Some  filenames containing the "%" character cannot be loaded by Maple.  E.g.  "%add.mw"  or "xx%abyy.mw".  But "xx%yy.mw" is OK.
This happens in Windows, but I suspect that it is not OS related.  It seems that Maple is trying to do some hex manipulations with the filename.

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