In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
to the quantum many-body hamiltonian
$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.
Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.
Can someone give a heuristic explanation also?
quantum-mechanics classical-mechanics operators hamiltonian representation-theory
edited 12 hours ago
Qmechanic♦
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
$endgroup$
add a comment |
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
to the quantum many-body hamiltonian
$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.
Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.
Can someone give a heuristic explanation also?
quantum-mechanics classical-mechanics operators hamiltonian representation-theory
edited 12 hours ago
Qmechanic♦
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
$endgroup$
3
$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
2
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago
add a comment |
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
to the quantum many-body hamiltonian
$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.
Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.
Can someone give a heuristic explanation also?
quantum-mechanics classical-mechanics operators hamiltonian representation-theory
edited 12 hours ago
Qmechanic♦
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
$endgroup$
When we go from the classical many-body hamiltonian
$$H = sum_i frac{vec{p}_i^2}{2m_e} - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} + sum_I frac{vec{p}_I^2}{2M_I}+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
to the quantum many-body hamiltonian
$$H = -sum_i frac{hbar^2}{2m_e}nabla_i^2 - sum_{i,I} frac{Z_I e^2 }{|vec{r}_i - vec{R}_I|} + frac{1}{2}sum_{i,j} frac{ e^2 }{|vec{r}_i - vec{r}_j|} - sum_I frac{hbar^2}{2M_I} nabla_I^2+ frac{1}{2}sum_{I,J} frac{Z_IZ_J e^2 }{|vec{R}_I - vec{R}_J|}$$
only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.
Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.
Can someone give a heuristic explanation also?
quantum-mechanics classical-mechanics operators hamiltonian representation-theory
quantum-mechanics classical-mechanics operators hamiltonian representation-theory
edited 12 hours ago
Qmechanic♦
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
edited 12 hours ago
Qmechanic♦
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
edited 12 hours ago
Qmechanic♦
106k121941217
edited 12 hours ago
Qmechanic♦
106k121941217
edited 12 hours ago
Qmechanic♦
106k121941217
106k121941217
asked 13 hours ago
UriAcevesUriAceves
788
asked 13 hours ago
UriAcevesUriAceves
788
asked 13 hours ago
UriAcevesUriAceves
788
788
3
$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
2
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago
add a comment |
3
$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
2
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago
3
3
$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
2
2
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.
Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.
In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
$endgroup$
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
add a comment |
$begingroup$
OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.
Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.
In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
$endgroup$
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
add a comment |
$begingroup$
Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.
Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.
In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
$endgroup$
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
add a comment |
$begingroup$
Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.
Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.
In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
$endgroup$
Actually the potential is also an operator. It just so happens that, in the position representation, $hat xpsi(x)=xpsi(x)$, so that the potential energy operator $V(hat x)$ acts by multiplication: $V(hat x)psi(x)=V(x)psi(x)$.
Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.
In the momentum representation $hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
answered 12 hours ago
ZeroTheHeroZeroTheHero
20.8k53261
20.8k53261
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
add a comment |
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
1
1
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
$begingroup$
Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure).
$endgroup$
– yuggib
10 hours ago
add a comment |
$begingroup$
OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
$endgroup$
add a comment |
$begingroup$
OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
$endgroup$
add a comment |
$begingroup$
OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
$endgroup$
OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
edited 11 hours ago
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
answered 12 hours ago
Qmechanic♦Qmechanic
106k121941217
106k121941217
add a comment |
add a comment |
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$begingroup$
The position operator (and its functions) are multiplication operators, in the standard Schrodinger position representation of canonical commutation relations. A multiplication operator is not acting as a number, but rather as the multiplication by a function. This is the "Fourier analogous" of a partial- or pseudo- differential operator, and in fact in momentum representation potentials are partial or pseudo differential operators.
$endgroup$
– yuggib
12 hours ago
2
$begingroup$
The potential operator is actually: $hat{V} equiv V(x)hat{I}$ with $hat{I}$ the identity operator for the 1D case here. Often you omit the $hat{I}$.
$endgroup$
– Dani
7 hours ago