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Transcript

Dirac Notation in Mathematica by José Luis Gómez-Muñoz http://homepage.cem.itesm.mx/lgomez/quantum/ [email protected] Introduction This is a tutorial on the use of Quantum Mathematica add-on to enter kets, bras and other quantum objects in Dirac notation. Load the Package First load the Quantum`Notation` package. Write: Needs["Quantum`Notation`"]; then press at the same time the keys ˜-Û to evaluate. Mathematica will load the package and print a welcome message: [email protected]"Quantum`Notation`"D Quantum`Notation` Version 2.2.0. HJuly 2010L A Mathematica package for Quantum calculations in Dirac bra−ket notation by José Luis Gómez−Muñoz Execute [email protected] in order to use the keyboard to enter quantum objects in Dirac's notation [email protected] must be executed again in each new notebook that is created, only one time per notebook. In order to use the keyboard to enter quantum objects write: SetQuantumAliases[ ]; then press at the same time the keys ˜-Û to evaluate. Remember that SetQuantumAliases[ ] must be evaluated again in each new notebook: 2 v7diracnotation.nb [email protected] ALIASES: @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] @[email protected] ket template bra template braket template operator template quantum concatenation infix symbol quantum concatenation infix symbol tensor product infix symbol quantum product template sigma notation for sums template sigma notation for sums template eigenvalue−label template eigenstate template two−operators−eigenstate template three−operators−eigentstate template bra of eigenstate template bra of two−operators−eigenstate template bra of three−operators−eigentstate template braket of eigenstates template braket of two−operators−eigenstates template braket of three−operators−eigentstate template operator HmatrixL element template operator HmatrixL element template operator HmatrixL element template operator HmatrixL element template hermitian conjugate template complex conjugate template quantum norm template partial trace template commutator template anticommutator template subscript template power template The quantum concatenation infix symbol @[email protected] is used for operator application, inner product and outer product. [email protected] must be executed again in each new notebook that is created, only one time per notebook. Entering Kets, Bras and Brakets In order to write a ket in Dirac's notation, place the cursor in a new Mathematica cell and press the keyboard keys: [ESC]ket[ESC] The ket template will appear. In order to select and fill in the "place holder" (square) press the keys: [TAB] z Finally press at the same time the keys[SHIFT] and[ENTER] to evaluate. z\ z\ v7diracnotation.nb In a similar way you can enter a bra: [ESC]bra[ESC] then press [TAB] and fill in the "place holder" (square) with label z: Xz Xz Here is a braket: [ESC]braket[ESC] [TAB]a[TAB]b Xa b\ Xa b\ The internal product of a bra and a ket is entered by pressing the keys: [ESC]bra[ESC] [ESC]on[ESC] [ESC]ket[ESC] press [TAB] one or two times to select the first "place holder" (square) and press: a[TAB]b finally press at the same time [SHIFT]-[ENTER] Xa ⋅ Xa b\ b\ Entering Kets of orthonormal states In order to write the eigenket of operator p with eigenvalue 3, place the cursor in a new Mathematica cell and press the keyboard keys: [ESC]eket[ESC] The eigenket template will appear. Next press the keys: [TAB]3[TAB]p Finally press at the same time the keys [SHIFT] and [ENTER] to evaluate: 3ˆp ] 3ˆp ] In order to write the operator h acting on its eigenket with eigenvalue 3 press the keys: [ESC]op[ESC] [ESC]on[ESC] [ESC]eket[ESC] The operator and eigenket templates will appear. Next press [TAB] several times till the first place holder (square) is selected. Then press the keys: h[TAB]3[TAB]h Finally press at the same time the keys [SHIFT] and [ENTER] to evaluate. The result of the calculation is the same ket multiplied by its eigenvalue: 3 4 v7diracnotation.nb ˆ h⋅ 3ˆh ] 3ˆh ] 3 In order to write a ket that is eigenket of two operators press the keys: [ESC]eeket[ESC] The corresponding eigenket template will appear. Next press the keys: [TAB]3[TAB]p[TAB]1[TAB]c Finally press at the same time the keys [SHIFT] and [ENTER] to evaluate. 3ˆp , 1ˆc ] 1ˆc , 3ˆp ] Another example of an operator acting on its eigenket: [ESC]op[ESC] [ESC]on[ESC] [ESC]eeeket[ESC] The operator and eigenket templates will appear. Next press [TAB] several times till the first place holder (square) is selected. Then press the keys: b[TAB]2[TAB]c[TAB]3[TAB]a[TAB]4[TAB]b Finally press at the same time the keys [SHIFT] and [ENTER] to evaluate. The result of the calculation is the same ket multiplied by its eigenvalue: ˆ b⋅ 4 2ˆc , 3ˆa , 4ˆb ] 3aˆ , 4ˆb , 2cˆ ] In order to enter the bra that corresponds to an eigenket press: [ESC]ebra[ESC] then press [TAB] and fill in the first "place holder" (square) with 6. Press [TAB] again and fill in the second "place holder" with m: X6mˆ X6mˆ Next calculation gives one because eigenstates of the same operator are assumed to be orthonormal: [ESC]ebra[ESC] [ESC]on[ESC] [ESC]eket[ESC] Next press [TAB] several times till the first place holder (square) is selected. Then press: 2[TAB]m[TAB]2[TAB]m X2mˆ ⋅ 2mˆ \ 1 Next calculation gives zero because eigenstates of the same operator are assumed to be orthonormal: v7diracnotation.nb X2mˆ ⋅ 3mˆ \ 0 This calculation gives a KroneckerDelta because eigenstates of the same operator are assumed to be orthonormal: Xamˆ ⋅ bmˆ \ [email protected] − bD Here is a braket made of eigenstates [ESC]eebraket[ESC] [TAB]2[TAB]q[TAB]3[TAB]c[TAB]2[TAB]q[TAB]3[TAB]c Y2ˆq , 3ˆc 2ˆq , 3ˆc ] 1 In order to write a tensor product of kets press the keys: [ESC]eket[ESC] [ESC]on[ESC] [ESC]eket[ESC] Next press[TAB] several times till the first place holder (square) is selected. Then press: 3[TAB]q[TAB]1[TAB]c 3ˆq ] ⋅ 1ˆc ] 1ˆc , 3ˆq ] Using tensor products you can generate a ket of any length: 2vˆ , 8ˆr , 9ˆt ] ⋅ 5ˆa , 6ˆc ] ⋅ 3ˆw , − 7ˆz , aˆf ] 5ˆa , 6ˆc , aˆf , 8ˆr , 9ˆt , 2ˆv , 3ˆw , H− 7Lˆz ] You can apply an operator in a subspace to a ket of the space: [ESC]eket[ESC] [ESC]on[ESC] [ESC]ebra[ESC] [ESC]on[ESC] [ESC]eeket[ESC] Next press[TAB] several times till the first place holder (square) is selected.Then press: a[TAB]q[TAB]j[TAB]q[TAB]0[TAB]c[TAB]j[TAB]q and press at the same time[SHIFT]-[ENTER] aˆq ] ⋅ Yjˆq 0ˆc , aˆq ] ⋅ 0ˆc , jˆq ] 5 6 v7diracnotation.nb Hermitian Conjugate In order to calculate the Hermitian Conjugate of a ket press the keys: [ESC]her[ESC] [TAB] [ESC]ket[ESC] [TAB]a and press at the same time [SHIFT]-[ENTER] H a\L† Xa In order to calculate the Hermitian Conjugate of an expresion press the following keys (Notice that the imaginary i is entered with two ii between two [ESC]): [ESC]her[ESC][TAB] (8+9[ESC]ii[ESC])*[ESC]eeket[ESC]+(5+7[ESC]ii[ESC])*[ESC]eeket[ESC] Next press [TAB] several times till the first place holder (square) is selected. Then press: 0[TAB]c[TAB]3[TAB]q[TAB]1[TAB]c[TAB]-5[TAB]q and press at the same time [SHIFT]-[ENTER] IH8 + 9 L 0ˆc , 3ˆq ] + H5 + 7 L ∗ H8 − 9 L Y0ˆc , 3ˆq 1ˆc , − 5ˆq ]M † + H5 − 7 L Y1ˆc , H− 5Lˆq It also works in symbolic expressions [ESC]her[ESC][TAB] a[SPACE][ESC]ket[ESC]+[ESC]con[ESC][SPACE][ESC]ket[ESC] Next press [TAB] several times till the first place holder (square) is selected. Then press: x[TAB]b[TAB]y and press at the same time [SHIFT]-[ENTER] Ha x\ + HbL∗ a∗ Xx y\L † + b Xy Superpositions of Kets and Operators This is the way to define a ket: [ESC]ket[ESC]=a[ESC]eket[ESC]+b[ESC]eket[ESC]+c[ESC]eket[ESC]+d[ESC]eket[ESC]+e[ESC]eket[ESC] Next press [TAB] several times till the first place holder (square) is selected. Then press: [ESC]psi[ESC][TAB]1[TAB]q[TAB]2[TAB]q[TAB]3[TAB]q[TAB]4[TAB]q[TAB]5[TAB]q finally press at the same time [SHIFT]-[ENTER] ψ\ = a a 1ˆq ] + b 1ˆq ] + b 2ˆq ] + c 2ˆq ] + c 3ˆq ] + d 3ˆq ] + d 4ˆq ] + e 4ˆq ] + e 5ˆq ] 5ˆq ] v7diracnotation.nb Mathematica calculates the corresponding bra: [ESC]bra[ESC] [TAB] [ESC]psi[ESC] finally press at the same time [SHIFT]-[ENTER] Xψ a∗ Y1ˆq + b∗ Y2ˆq + c∗ Y3ˆq + d∗ Y4ˆq + e∗ Y5ˆq The internal product of a ket with its dual is a real number (square of its norm) [ESC]bra[ESC] [ESC]on[ESC] [ESC]ket[ESC] press [TAB] one or two times to select the first "place holder" (square) and press: [ESC]psi[ESC] [TAB] [ESC]psi[ESC] finally press at the same time [SHIFT]-[ENTER] Xψ ⋅ ψ\ a a∗ + b b∗ + c c∗ + d d∗ + e e∗ Mathematica calculates the corresponding norm: [ESC]norm[ESC] [TAB] [ESC]ket[ESC] [TAB] [ESC]psi[ESC] finally press at the same time [SHIFT]-[ENTER] ∞ ψ\¥ a a∗ + b b∗ + c c∗ + d d∗ + e e∗ This external product is not immediately calculated: ψ\ ⋅ Xψ Ia 1ˆq ] + b ∗ Ia Y1qˆ 2ˆq ] + c ∗ + b Y2qˆ 3ˆq ] + d ∗ + c Y3qˆ 4ˆq ] + e ∗ + d Y4qˆ 5ˆq ]M ⋅ ∗ + e Y5qˆ M The command Expand[] can be used to calculate the external product [email protected] a a∗ da ψ\ ⋅ Xψ D 1ˆq ] ⋅ Y1ˆq ∗ 4ˆq ] ⋅ Y1ˆq + b a∗ +ea 2ˆq ] ⋅ Y1ˆq ∗ 5ˆq ] ⋅ Y1ˆq + c a∗ +ab 3ˆq ] ⋅ Y1ˆq ∗ 1ˆq ] ⋅ + Y2ˆq + b b∗ 2ˆq ] ⋅ Y2ˆq + c b∗ 3ˆq ] ⋅ Y2ˆq + d b∗ 4ˆq ] ⋅ Y2ˆq + e b∗ 5ˆq ] ⋅ Y2ˆq + a c∗ 1ˆq ] ⋅ Y3ˆq + b c∗ 2ˆq ] ⋅ Y3ˆq + c c∗ 3ˆq ] ⋅ Y3ˆq + d c∗ 4ˆq ] ⋅ Y3ˆq + e c∗ 5ˆq ] ⋅ Y3ˆq + a d∗ 1ˆq ] ⋅ Y4ˆq + b d∗ 2ˆq ] ⋅ Y4ˆq + c d∗ 3ˆq ] ⋅ Y4ˆq + d d∗ 4ˆq ] ⋅ Y4ˆq + e d∗ 5ˆq ] ⋅ Y4ˆq + a e∗ 1ˆq ] ⋅ Y5ˆq + 2ˆq ] ⋅ Y5ˆq ∗ 3ˆq ] ⋅ Y5ˆq ∗ 4ˆq ] ⋅ Y5ˆq ∗ 5ˆq ] ⋅ Y5ˆq be ∗ +ce +de +ee 7 8 v7diracnotation.nb Here we define another ket [ESC]ket[ESC]=u[ESC]eket[ESC]+v[ESC]eket[ESC]+w[ESC]eket[ESC]+x[ESC]eket[ESC]+y[ESC]eket[ESC] Next press [TAB] several times till the first place holder (square) is selected. Then press: m[TAB]1[TAB]q[TAB]2[TAB]q[TAB]3[TAB]q[TAB]4[TAB]q[TAB]5[TAB]q finally press at the same time [SHIFT]-[ENTER] 1ˆq ] + v m\ = u 1ˆq ] + v u 2ˆq ] + w 2ˆq ] + w 3ˆq ] + x 3ˆq ] + x 4ˆq ] + y 4ˆq ] + y 5ˆq ] 5ˆq ] Note for advanced Mathematica users: The definition is stored as an upvalue of the variable m: ?m Global`m 1ˆq ] + v m\ ^= u 2ˆq ] + w 3ˆq ] + x 4ˆq ] + y 5ˆq ] Here is an operator made of the states that were defined before: [email protected] a u∗ ψ\ ⋅ Xm D + b u∗ 1ˆq ] ⋅ Y1ˆq du ∗ bv ∗ ev ∗ 5qˆ ] ⋅ Y2qˆ dw ∗ 4ˆq ] cx ∗ 3ˆq ] by ∗ 2ˆq ] 4ˆq ] ⋅ Y1ˆq 2ˆq ] ⋅ Y2ˆq ⋅ Y3ˆq ⋅ Y4ˆq ⋅ Y5ˆq 2ˆq ] ⋅ Y1ˆq +eu ∗ +cv ∗ +aw ∗ 1qˆ ] ⋅ Y3qˆ +ew ∗ 5ˆq ] +dx ∗ 4ˆq ] +cy ∗ 3ˆq ] 5ˆq ] ⋅ Y1ˆq 3ˆq ] ⋅ Y2ˆq ⋅ Y3ˆq ⋅ Y4ˆq ⋅ Y5ˆq + c u∗ 3ˆq ] ⋅ Y1ˆq + +av ∗ 1ˆq ] ⋅ Y2ˆq + +dv ∗ 4ˆq ] ⋅ Y2ˆq + +bw ∗ 2qˆ ] ⋅ Y3qˆ + c w∗ +ax ∗ 1ˆq ] +ex ∗ 5ˆq ] +dy ∗ 4ˆq ] ⋅ Y4ˆq ⋅ Y4ˆq ⋅ Y5ˆq 3qˆ ] ⋅ Y3qˆ + +bx ∗ 2ˆq ] ⋅ Y4ˆq + +ay ∗ 1ˆq ] ⋅ Y5ˆq + +ey ∗ 5ˆq ] ⋅ Y5ˆq ˆ We can obtain the partial trace of the operator. The base-operator template is entered [ESC]op[ESC]: QuantumPartialTraceA ˆ ψ\ ⋅ Xm , qE a u∗ + b v∗ + c w∗ + d x∗ + e y∗ Here we apply the "base" operator q to the ket. Notice that q is inside the operator template [ESC]op[ESC] ˆ q⋅ a ψ\ 1ˆq ] + 2 b 2ˆq ] + 3 c 3ˆq ] + 4 d 4ˆq ] + 5 e 5ˆq ] This is one way to define another operator. Notice that p is not inside the operator template: v7diracnotation.nb 1qˆ ] ⋅ Y2qˆ p=f f 1ˆq ] ⋅ Y2ˆq 2qˆ ] ⋅ Y4qˆ +g 2ˆq ] ⋅ Y4ˆq +g Now the operator and ket that were defined can be used. Notice that p is not inside the operator template: p⋅ ψ\ bf 1ˆq ] + d g 2ˆq ] Notice that p is not inside the operator template, but base operator q is inside the template: ˆ p⋅q⋅ ψ\ 1ˆq ] + 4 d g 2bf 2ˆq ] Hermitian conjugate [ESC]her[ESC] HpL† f∗ 2ˆq ] ⋅ Y1ˆq + g∗ 4ˆq ] ⋅ Y2ˆq An expression involving a bra, an operator and a ket: Xψ ⋅p⋅ ψ\ b f a∗ + d g b∗ Another expression involving a bra, an operator and a ket: Xψ ˆ ⋅q⋅ ψ\ a a∗ + 2 b b∗ + 3 c c∗ + 4 d d∗ + 5 e e∗ Here is another operator : ope = a a 1ˆq , 3ˆs ] ⋅ Y1ˆq , 3ˆs 1qˆ , 3sˆ ] ⋅ Y1qˆ , 3sˆ +b +b 2ˆq , 4ˆs ] ⋅ Y3ˆq , 2ˆs 2qˆ , 4sˆ ] ⋅ Y3qˆ , 2sˆ +c +c 3ˆq , 5ˆs ] ⋅ Y4ˆq , 5ˆs 3qˆ , 5sˆ ] ⋅ Y4qˆ , 5sˆ ˆ Mathematica can calculate the partial trace with respect to operator q: [ESC]trace[ESC] 9 10 v7diracnotation.nb Trˆq @opeD a 3ˆs ] ⋅ Y3ˆs ˆ Mathematica can calculate the partial trace with respect to operator s: [ESC]trace[ESC] Trˆs @opeD a 1ˆq ] ⋅ Y1ˆq +c 3ˆq ] ⋅ Y4ˆq Undefined Symbols are Assumed to be Scalars Any undefined name, like B, is assumed to be a complex scalar: [ESC]her[ESC] [email protected]; HXα ⋅ B ⋅ B∗ Xβ β\L† α\ SetQuantumObject[B] specifies that B is not a complex scalar: [email protected] The object B will Not be considered as a complex scalar The object [email protected] will Not be considered as a complex scalar The object [email protected], __D will Not be considered as a complex scalar The object [email protected], [email protected] will Not be considered as a complex scalar The object [email protected]@___D, __D will Not be considered as a complex scalar The object [email protected]@___D, [email protected] will Not be considered as a complex scalar After executing SetQuantumObject[B], B is considered an operator: HXα Xβ ⋅B⋅ ⋅ B† ⋅ β\L† α\ Finally we clear the definitions made in this document: [email protected], p, ope, ψ, BD v7diracnotation.nb 11 ReplaceAll versus QuantumReplaceAll ReplaceAll is a standard Mathematica command that can take advantage of the pattern recognitizion language of Mathematica. The "delayed rule" symbol ß can be entered by pressing the keys [ESC]:>[ESC] φ\ + b ReplaceAllBa aH 0\ + 1\L +b ψ\, φ\ H 0\ + 1\L í 2 F ψ\ 2 However, ReplaceAll will not make any replacement in the following case, because the expression φ\ ⊗ ψ\ evolves to φ, ψ\, and the Mathematica command ReplaceAll does not recognize the ket φ\ in this evolved expression: φ\ ⊗ ReplaceAllB ψ\, φ\ H 0\ + 1\L í 2 F φ, ψ\ On the other hand, the Quantum Mathematica command QuantumReplace does recognize the ket and performs the replacement: QuantumReplaceB 0, ψ\ + φ\ ⊗ ψ\, φ\ H 0\ + 1\L í 2 F 1, ψ\ 2 QuantumReplaceAll will also work on bras: QuantumReplaceB ψ\ ⋅ HX0 ψ\ ⋅ Xφ , φ\ H 0\ + 1\L í 2 F + X1 L 2 ExpandBQuantumReplaceB ψ\ ⋅ X0 2 + ψ\ ⋅ X1 2 ψ\ ⋅ Xφ , φ\ H 0\ + 1\L í 2 FF 12 v7diracnotation.nb Questions and Exercises 1. What command is used to load the Quantum Notation package in a fresh Mathematica session? Answer: 2. What command is used to load the Dirac Notation's keyboard aliases in a new Mathematica document (notebook)? Answer: 3. Select the correct sentence. There is only one: a) SetQuantumAliases[] must be executed after evaluating Needs["Quantum`Notation`"] in a fresh Mathematica session b) SetQuantumAliases[] can be executed before evaluating Needs["Quantum`Notation`"] in a fresh Mathematica session c) SetQuantumAliases[] must be executed before evaluating Needs["Quantum`Notation`"] in a fresh Mathematica session The correct sentence is:_____ 4. Select the correct sentence. There is only one: a) SetQuantumAliases[] must be evaluated before executing each command in Mathematica b) SetQuantumAliases[] must be evaluated in each new document (notebook) in Mathematica c) SetQuantumAliases[] must be evaluated in each fresh session in Mathematica The correct sentence is:_____ 5. What combination of keys (keyboard alias) must be pressed in order to obtain the "CenterDot" Answer: 6. Select the correct sentences for the Quantum Package. There is more than one correct sentence: a) CenterDot · represents the internal product of a bra and a ket b) CenterDot · represents the hermitian conjugate operation c) CenterDot · represents the external product of a ket and a bra d) CenterDot · represents the partial trace operation e) CenterDot · represents the application of an operator to a ket The correct sentences are:_____ 7. What is the difference between 3\ y 3ˆA ] in the Quantum Mathematica Package? Answer: 8. What differences are there between the Dirac notation as used in Quantum Mechanics textbooks and the notation used in the Quantum Mathematica package? Answer: v7diracnotation.nb 13 9. What similarities are there between the Dirac notation as used in Quantum Mechanics textbooks and the notation used in the Quantum Mathematica package? Answer: 10. Why the following command generates an error message? 3\ = a\ + b\ Why the following command does not generate the same error message? m\ = a\ + b\ Answer: by José Luis Gómez-Muñoz http://homepage.cem.itesm.mx/lgomez/quantum/ [email protected]