In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?Notation in a paper on quantum mechanics and gravitation (2)System of two harmonic oscillators and its quantum partition functionWhy does time evolution operator have the form $U(t) = e^-itH$?Many-particle operators in the occupation number representationThe central field approximation and the quantum number $n$Can we always express the EM-Field Hamiltonian as (possibly time dependent) pair of annihilation and creation operators?Do we chose an operator ordering (explicit or implicit) all the time?Dimensions in the Second Quantization of an OperatorQuantum numbers $n$ and $ell$ relation in many electrons atomWhy do we think that the relation $vecmu_L=frace2m_evecL$ will be valid in quantum mechanics?
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In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?
Notation in a paper on quantum mechanics and gravitation (2)System of two harmonic oscillators and its quantum partition functionWhy does time evolution operator have the form $U(t) = e^-itH$?Many-particle operators in the occupation number representationThe central field approximation and the quantum number $n$Can we always express the EM-Field Hamiltonian as (possibly time dependent) pair of annihilation and creation operators?Do we chose an operator ordering (explicit or implicit) all the time?Dimensions in the Second Quantization of an OperatorQuantum numbers $n$ and $ell$ relation in many electrons atomWhy do we think that the relation $vecmu_L=frace2m_evecL$ will be valid in quantum mechanics?
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i fracvecp_i^22m_e - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 + sum_I fracvecp_I^22M_I+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
to the quantum many-body hamiltonian
$$H = -sum_i frachbar^22m_enabla_i^2 - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 - sum_I frachbar^22M_I nabla_I^2+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
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♦
106k121941220
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
add a comment |
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i fracvecp_i^22m_e - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 + sum_I fracvecp_I^22M_I+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
to the quantum many-body hamiltonian
$$H = -sum_i frachbar^22m_enabla_i^2 - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 - sum_I frachbar^22M_I nabla_I^2+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
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♦
106k121941220
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
2
$begingroup$
The potential operator is actually: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– Dani
Mar 11 at 14:57
add a comment |
$begingroup$
When we go from the classical many-body hamiltonian
$$H = sum_i fracvecp_i^22m_e - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 + sum_I fracvecp_I^22M_I+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
to the quantum many-body hamiltonian
$$H = -sum_i frachbar^22m_enabla_i^2 - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 - sum_I frachbar^22M_I nabla_I^2+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
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♦
106k121941220
asked Mar 11 at 9:00
UriAcevesUriAceves
858
$endgroup$
When we go from the classical many-body hamiltonian
$$H = sum_i fracvecp_i^22m_e - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 + sum_I fracvecp_I^22M_I+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
to the quantum many-body hamiltonian
$$H = -sum_i frachbar^22m_enabla_i^2 - sum_i,I fracZ_I e^2 + frac12sum_i,j frac e^2 - sum_I frachbar^22M_I nabla_I^2+ frac12sum_I,J fracZ_IZ_J e^2 vecR_I - vecR_J$$
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♦
106k121941220
asked Mar 11 at 9:00
UriAcevesUriAceves
858
edited Mar 11 at 10:10
Qmechanic♦
106k121941220
asked Mar 11 at 9:00
UriAcevesUriAceves
858
edited Mar 11 at 10:10
Qmechanic♦
106k121941220
edited Mar 11 at 10:10
Qmechanic♦
106k121941220
edited Mar 11 at 10:10
Qmechanic♦
106k121941220
106k121941220
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: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– Dani
Mar 11 at 14:57
add a comment |
2
$begingroup$
The potential operator is actually: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– Dani
Mar 11 at 14:57
2
2
$begingroup$
The potential operator is actually: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– Dani
Mar 11 at 14:57
$begingroup$
The potential operator is actually: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– 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
20.9k53361
$endgroup$
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
106k121941220
$endgroup$
add a comment |
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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 Mar 11 at 10:06
ZeroTheHeroZeroTheHero
20.9k53361
$endgroup$
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
20.9k53361
$endgroup$
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
20.9k53361
$endgroup$
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
20.9k53361
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
20.9k53361
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
20.9k53361
answered Mar 11 at 10:06
ZeroTheHeroZeroTheHero
20.9k53361
20.9k53361
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
106k121941220
$endgroup$
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
106k121941220
$endgroup$
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
106k121941220
$endgroup$
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
106k121941220
edited Mar 11 at 11:14
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121941220
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121941220
answered Mar 11 at 10:07
Qmechanic♦Qmechanic
106k121941220
106k121941220
add a comment |
add a comment |
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$begingroup$
The potential operator is actually: $hatV equiv V(x)hatI$ with $hatI$ the identity operator for the 1D case here. Often you omit the $hatI$.
$endgroup$
– Dani
Mar 11 at 14:57