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 Mar 11 at 10:10
Qmechanic♦
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
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 Mar 11 at 10:10
Qmechanic♦
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
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
Mar 11 at 14:57
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 Mar 11 at 10:10
Qmechanic♦
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
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 Mar 11 at 10:10
Qmechanic♦
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
edited Mar 11 at 10:10
Qmechanic♦
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
edited Mar 11 at 10:10
Qmechanic♦
106k121961223
edited Mar 11 at 10:10
Qmechanic♦
106k121961223
edited Mar 11 at 10:10
Qmechanic♦
106k121961223
106k121961223
asked Mar 11 at 9:00
UriAcevesUriAceves
858
asked Mar 11 at 9:00
UriAcevesUriAceves
858
asked Mar 11 at 9:00
UriAcevesUriAceves
858
858
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
Mar 11 at 14:57
add a comment |
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
Mar 11 at 14:57
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
Mar 11 at 14:57
$begingroup$
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
Mar 11 at 14:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
$endgroup$
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
Mar 11 at 12:30
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 Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
$endgroup$
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
Mar 11 at 12:30
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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
$endgroup$
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
Mar 11 at 12:30
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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
$endgroup$
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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
21k53363
21k53363
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
Mar 11 at 12:30
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
Mar 11 at 12:30
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
Mar 11 at 12:30
$begingroup$
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
Mar 11 at 12:30
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 Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
$endgroup$
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 Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
$endgroup$
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 Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
$endgroup$
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 Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
edited Mar 11 at 11:14
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121961223
106k121961223
add a comment |
add a comment |
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$begingroup$
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
Mar 11 at 14:57