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Writing operators in the position basis
What is the physical interpretation of the density matrix in a double continuous basis $|alpharangle$, $|betarangle$?Matrix elements of the operator $hatx hatp$ in position and momentum basisCommutation relation of position and momentum using Dirac notationWhat's the correct link between Dirac notation and wave mechanics integrals?Definitions of position operator in QMMomentum operator in position basisCorrespondence between integral transformations and differential operators in quantum mechanicsTaking the Hamiltonian eigenvalue problem into position space?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
To calculate an expecation value in quantum mechanics of some operator $hat A$ you can generally write it in Dirac notation
$$langle hat Arangle=langlepsi|hat A|psirangle.$$
In the position basis this becomes the following integral:
$$int mathrm dx psi^*(x)A(x)psi(x).$$
But when I try to derive this explicitly from Dirac notation I get
beginalignlanglepsi|hat A|psirangle
&=langlepsi|left(int mathrmdx|xranglelangle x|right)hat Aleft(int mathrmdx'|x'ranglelangle x'|right)|psirangle\
&=int mathrmdx mathrmdx'langlepsi|xranglelangle x|hat A|x'ranglelangle x|psirangle\
&=int mathrmdx mathrmdx' psi^*(x),langle x|hat A|x'rangle,psi(x').
endalign
The reason I'm stuck is that I don't remember ever calculating the matrix elements in the position basis. How do you calculate these matrix elements in the position basis (or more generally in any basis)?
quantum-mechanics hilbert-space operators wavefunction
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
To calculate an expecation value in quantum mechanics of some operator $hat A$ you can generally write it in Dirac notation
$$langle hat Arangle=langlepsi|hat A|psirangle.$$
In the position basis this becomes the following integral:
$$int mathrm dx psi^*(x)A(x)psi(x).$$
But when I try to derive this explicitly from Dirac notation I get
beginalignlanglepsi|hat A|psirangle
&=langlepsi|left(int mathrmdx|xranglelangle x|right)hat Aleft(int mathrmdx'|x'ranglelangle x'|right)|psirangle\
&=int mathrmdx mathrmdx'langlepsi|xranglelangle x|hat A|x'ranglelangle x|psirangle\
&=int mathrmdx mathrmdx' psi^*(x),langle x|hat A|x'rangle,psi(x').
endalign
The reason I'm stuck is that I don't remember ever calculating the matrix elements in the position basis. How do you calculate these matrix elements in the position basis (or more generally in any basis)?
quantum-mechanics hilbert-space operators wavefunction
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
To calculate an expecation value in quantum mechanics of some operator $hat A$ you can generally write it in Dirac notation
$$langle hat Arangle=langlepsi|hat A|psirangle.$$
In the position basis this becomes the following integral:
$$int mathrm dx psi^*(x)A(x)psi(x).$$
But when I try to derive this explicitly from Dirac notation I get
beginalignlanglepsi|hat A|psirangle
&=langlepsi|left(int mathrmdx|xranglelangle x|right)hat Aleft(int mathrmdx'|x'ranglelangle x'|right)|psirangle\
&=int mathrmdx mathrmdx'langlepsi|xranglelangle x|hat A|x'ranglelangle x|psirangle\
&=int mathrmdx mathrmdx' psi^*(x),langle x|hat A|x'rangle,psi(x').
endalign
The reason I'm stuck is that I don't remember ever calculating the matrix elements in the position basis. How do you calculate these matrix elements in the position basis (or more generally in any basis)?
quantum-mechanics hilbert-space operators wavefunction
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
$endgroup$
To calculate an expecation value in quantum mechanics of some operator $hat A$ you can generally write it in Dirac notation
$$langle hat Arangle=langlepsi|hat A|psirangle.$$
In the position basis this becomes the following integral:
$$int mathrm dx psi^*(x)A(x)psi(x).$$
But when I try to derive this explicitly from Dirac notation I get
beginalignlanglepsi|hat A|psirangle
&=langlepsi|left(int mathrmdx|xranglelangle x|right)hat Aleft(int mathrmdx'|x'ranglelangle x'|right)|psirangle\
&=int mathrmdx mathrmdx'langlepsi|xranglelangle x|hat A|x'ranglelangle x|psirangle\
&=int mathrmdx mathrmdx' psi^*(x),langle x|hat A|x'rangle,psi(x').
endalign
The reason I'm stuck is that I don't remember ever calculating the matrix elements in the position basis. How do you calculate these matrix elements in the position basis (or more generally in any basis)?
quantum-mechanics hilbert-space operators wavefunction
quantum-mechanics hilbert-space operators wavefunction
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
edited Sep 30 at 16:05
Qmechanic♦
118k14 gold badges238 silver badges1409 bronze badges
118k14 gold badges238 silver badges1409 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
asked Sep 29 at 23:03
AccidentalTaylorExpansionAccidentalTaylorExpansion
2,68711 silver badges21 bronze badges
2,68711 silver badges21 bronze badges
add a comment
|
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (which can be provided directly, or in shorthand using the operator's position-space representation described below).
If $hatA$ happens to have every $|xrangle$ basis state as an eigenstate, then you can write $hatA|xrangle=A(x)|xrangle$, where $A(x)$ is the operator's position-basis representation. Then you can say that $langle x'|hatA|xrangle=A(x)langle x'|xrangle=A(x)delta(x-x')$. For example, for the simple-harmonic-oscillator potential operator $hatA=kappahatx^2$, we have that $langle x'|hatA|xrangle=kappa x^2delta(x-x')$.
This condition is not necessarily true in general, though, and there are many operators which simply don't have a nice position-space representation like this. My favorite example is the position translation operator $hatT_a$. The action of this operator on the basis states is defined to be as follows: $hatT_a|xrangle=|x+arangle$. As you can see, this operator has none of the position basis states as eigenstates! So the matrix element will look slightly different:
$$langle x'|hatT_a|xrangle=langle x'|x+arangle=delta(x+a-x')$$
which you'll notice doesn't conform to the structure $A(x)delta(x-x')$.
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
add a comment
|
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1 Answer
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1 Answer
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$begingroup$
In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (which can be provided directly, or in shorthand using the operator's position-space representation described below).
If $hatA$ happens to have every $|xrangle$ basis state as an eigenstate, then you can write $hatA|xrangle=A(x)|xrangle$, where $A(x)$ is the operator's position-basis representation. Then you can say that $langle x'|hatA|xrangle=A(x)langle x'|xrangle=A(x)delta(x-x')$. For example, for the simple-harmonic-oscillator potential operator $hatA=kappahatx^2$, we have that $langle x'|hatA|xrangle=kappa x^2delta(x-x')$.
This condition is not necessarily true in general, though, and there are many operators which simply don't have a nice position-space representation like this. My favorite example is the position translation operator $hatT_a$. The action of this operator on the basis states is defined to be as follows: $hatT_a|xrangle=|x+arangle$. As you can see, this operator has none of the position basis states as eigenstates! So the matrix element will look slightly different:
$$langle x'|hatT_a|xrangle=langle x'|x+arangle=delta(x+a-x')$$
which you'll notice doesn't conform to the structure $A(x)delta(x-x')$.
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
add a comment
|
$begingroup$
In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (which can be provided directly, or in shorthand using the operator's position-space representation described below).
If $hatA$ happens to have every $|xrangle$ basis state as an eigenstate, then you can write $hatA|xrangle=A(x)|xrangle$, where $A(x)$ is the operator's position-basis representation. Then you can say that $langle x'|hatA|xrangle=A(x)langle x'|xrangle=A(x)delta(x-x')$. For example, for the simple-harmonic-oscillator potential operator $hatA=kappahatx^2$, we have that $langle x'|hatA|xrangle=kappa x^2delta(x-x')$.
This condition is not necessarily true in general, though, and there are many operators which simply don't have a nice position-space representation like this. My favorite example is the position translation operator $hatT_a$. The action of this operator on the basis states is defined to be as follows: $hatT_a|xrangle=|x+arangle$. As you can see, this operator has none of the position basis states as eigenstates! So the matrix element will look slightly different:
$$langle x'|hatT_a|xrangle=langle x'|x+arangle=delta(x+a-x')$$
which you'll notice doesn't conform to the structure $A(x)delta(x-x')$.
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
add a comment
|
$begingroup$
In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (which can be provided directly, or in shorthand using the operator's position-space representation described below).
If $hatA$ happens to have every $|xrangle$ basis state as an eigenstate, then you can write $hatA|xrangle=A(x)|xrangle$, where $A(x)$ is the operator's position-basis representation. Then you can say that $langle x'|hatA|xrangle=A(x)langle x'|xrangle=A(x)delta(x-x')$. For example, for the simple-harmonic-oscillator potential operator $hatA=kappahatx^2$, we have that $langle x'|hatA|xrangle=kappa x^2delta(x-x')$.
This condition is not necessarily true in general, though, and there are many operators which simply don't have a nice position-space representation like this. My favorite example is the position translation operator $hatT_a$. The action of this operator on the basis states is defined to be as follows: $hatT_a|xrangle=|x+arangle$. As you can see, this operator has none of the position basis states as eigenstates! So the matrix element will look slightly different:
$$langle x'|hatT_a|xrangle=langle x'|x+arangle=delta(x+a-x')$$
which you'll notice doesn't conform to the structure $A(x)delta(x-x')$.
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
$endgroup$
In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (which can be provided directly, or in shorthand using the operator's position-space representation described below).
If $hatA$ happens to have every $|xrangle$ basis state as an eigenstate, then you can write $hatA|xrangle=A(x)|xrangle$, where $A(x)$ is the operator's position-basis representation. Then you can say that $langle x'|hatA|xrangle=A(x)langle x'|xrangle=A(x)delta(x-x')$. For example, for the simple-harmonic-oscillator potential operator $hatA=kappahatx^2$, we have that $langle x'|hatA|xrangle=kappa x^2delta(x-x')$.
This condition is not necessarily true in general, though, and there are many operators which simply don't have a nice position-space representation like this. My favorite example is the position translation operator $hatT_a$. The action of this operator on the basis states is defined to be as follows: $hatT_a|xrangle=|x+arangle$. As you can see, this operator has none of the position basis states as eigenstates! So the matrix element will look slightly different:
$$langle x'|hatT_a|xrangle=langle x'|x+arangle=delta(x+a-x')$$
which you'll notice doesn't conform to the structure $A(x)delta(x-x')$.
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 23:26
probably_someoneprobably_someone
25.2k1 gold badge37 silver badges76 bronze badges
25.2k1 gold badge37 silver badges76 bronze badges
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
add a comment
|
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
You are absolutely right. I deleted my answer.
$endgroup$
– Dani
Sep 29 at 23:34
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
Does that mean that $langlehat Arangle=int mathrm dx psi^*(x)A(x)psi(x)$ is not generally true? Would this be true for the (usual) Hamoltonian operator $hat H=hat p^2/(2m)+hat V$?
$endgroup$
– AccidentalTaylorExpansion
Sep 30 at 11:54
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
$begingroup$
@user3502079 It depends on whether or not you consider $A(x)$ to be a function (which yields a number) or a functional (a function which acts on other functions, for example $A(x)=fracpartialpartial x$). If $A(x)$ is a function, then no, you have $langle hatp^2rangle=int dx psi^*(x)hbar^2fracpartial^2psipartial x^2(x)$. If $A(x)$ is a functional, then you can declare $A(x)=hbar^2fracpartial^2partial x^2$.
$endgroup$
– probably_someone
Sep 30 at 14:08
add a comment
|
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