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Equivalent definitions of total angular momentum
Total angular momentum - single electronPhoton Angular MomentumAngular momentum of the electric field of a point-like electric charge and the magnetic field of a monopoleCan spin angular momentum be understood as orbital angular momentum in extra dimensions?Total angular momentum of a continuous bodyHow to count total spin degeneracies for many spin one half particles?Is Wikipedia's definition of angular velocity incorrect?Orbital angular momentum eigenstates in the $|mathbfrrangle$ representationDerivation of Angular Momentum OperatorDomain of symmetric momentum operator vs self-adjoint momentum operator
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Consider the equality
beginequationexpleft(-fracihbarboldsymbolphi Jright)left|xright>=left|R(phi)xright>,endequation
where $left|xright>$ denotes a position eigenstate, $J$ the total angular momentum operator on the ket space, and $R(boldsymbolphi)$ the $mathbbR^3$ rotation matrix for rotations around $boldsymbolphi=phicdotmathbfn$ with $||mathbfn||=1$.
Some consider this equation to be the definition of total angular momentum. How can this equation be proven by using the arguably more popular and classically motivated definition?
beginequation
mathbfJ:=mathbfL+mathbfS=mathbfXtimesmathbfP+mathbfS
endequation
Note: Thanks to the work of @Valter Moretti and @Adam Latosiński an equality between the two defintions has been established for a spinless particle.
quantum-mechanics angular-momentum operators hilbert-space definition
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
Consider the equality
beginequationexpleft(-fracihbarboldsymbolphi Jright)left|xright>=left|R(phi)xright>,endequation
where $left|xright>$ denotes a position eigenstate, $J$ the total angular momentum operator on the ket space, and $R(boldsymbolphi)$ the $mathbbR^3$ rotation matrix for rotations around $boldsymbolphi=phicdotmathbfn$ with $||mathbfn||=1$.
Some consider this equation to be the definition of total angular momentum. How can this equation be proven by using the arguably more popular and classically motivated definition?
beginequation
mathbfJ:=mathbfL+mathbfS=mathbfXtimesmathbfP+mathbfS
endequation
Note: Thanks to the work of @Valter Moretti and @Adam Latosiński an equality between the two defintions has been established for a spinless particle.
quantum-mechanics angular-momentum operators hilbert-space definition
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
$endgroup$
1
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
3
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
1
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05
|
show 1 more comment
$begingroup$
Consider the equality
beginequationexpleft(-fracihbarboldsymbolphi Jright)left|xright>=left|R(phi)xright>,endequation
where $left|xright>$ denotes a position eigenstate, $J$ the total angular momentum operator on the ket space, and $R(boldsymbolphi)$ the $mathbbR^3$ rotation matrix for rotations around $boldsymbolphi=phicdotmathbfn$ with $||mathbfn||=1$.
Some consider this equation to be the definition of total angular momentum. How can this equation be proven by using the arguably more popular and classically motivated definition?
beginequation
mathbfJ:=mathbfL+mathbfS=mathbfXtimesmathbfP+mathbfS
endequation
Note: Thanks to the work of @Valter Moretti and @Adam Latosiński an equality between the two defintions has been established for a spinless particle.
quantum-mechanics angular-momentum operators hilbert-space definition
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
$endgroup$
Consider the equality
beginequationexpleft(-fracihbarboldsymbolphi Jright)left|xright>=left|R(phi)xright>,endequation
where $left|xright>$ denotes a position eigenstate, $J$ the total angular momentum operator on the ket space, and $R(boldsymbolphi)$ the $mathbbR^3$ rotation matrix for rotations around $boldsymbolphi=phicdotmathbfn$ with $||mathbfn||=1$.
Some consider this equation to be the definition of total angular momentum. How can this equation be proven by using the arguably more popular and classically motivated definition?
beginequation
mathbfJ:=mathbfL+mathbfS=mathbfXtimesmathbfP+mathbfS
endequation
Note: Thanks to the work of @Valter Moretti and @Adam Latosiński an equality between the two defintions has been established for a spinless particle.
quantum-mechanics angular-momentum operators hilbert-space definition
quantum-mechanics angular-momentum operators hilbert-space definition
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
edited Jul 14 at 17:15
Qmechanic♦
111k12 gold badges214 silver badges1315 bronze badges
111k12 gold badges214 silver badges1315 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
asked Jul 14 at 5:32
TheoreticalMinimumTheoreticalMinimum
3491 silver badge12 bronze badges
3491 silver badge12 bronze badges
1
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
3
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
1
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05
|
show 1 more comment
1
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
3
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
1
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05
1
1
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
3
3
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
1
1
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Consider a smooth function $psi$ in $L^2(mathbbR^3, d^3x)$ (more precisely $psi$ is assumed to belong to Schwartz' space) and define for a fixed $n in mathbbS^2$ and $phi in mathbbR$
$$psi_phi(x)= psi(R^-1_n(phi)x)$$
Since $R^-1_n(phi) = e^-phi n cdot S$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$:
$$(S_j)_rs= epsilon_jrs$$
we have
$$fracddphipsi_phi(x) = -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi_phi(x):.tag1$$
Now consider
$$psi'_phi(x)= left(e^fracihbarphi n cdot Jpsiright)(x):,$$
where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only,
$$J_j = sum_r,s=1^3epsilon_jrs X_r P_s = -ihbar sum_r,s=1^3epsilon_jrs X_r fracpartialpartial x_s $$
Computing the $phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $psi$ matters), we have
$$fracddphipsi'_phi(x) = fracihbarsum_j=1^3 n cdot left(J_jpsi'_phiright)(x)= -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi'_phi(x):.tag2$$
In summary, for $x$ fixed, $psi_phi(x)$ and $psi'_phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition
$$psi_0(x)=psi'_0 (x):.$$
The uniqueness theorem for the solutions of first-order differential equations implies that
$$psi_phi(x)=psi'_phi (x):.$$
In other words
$$left(e^-fracihbar phi n cdot Jpsiright)(x) = psileft(R_n^-1(phi)xright):.$$
(The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.)
The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|yrangle = delta(x-y)= delta_y(x)$, so that, using rotational invariance of the Dirac delta function,
$$left(e^-fracihbar phi n cdot Jdelta_yright)(x) = delta_yleft(R_n^-1(phi)xright)= deltaleft(R_n^-1(phi)x -yright)= deltaleft(x -R_n(phi)yright):.$$
Coming back to the abstract notation, the found identity reads
$$e^-fracihbar phi n cdot J|yrangle = |R_n(phi)yrangle:.$$
(A completely rigorous proof can be found in this book of mine.)
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
$endgroup$
add a comment |
$begingroup$
The most broad definition of the angular momentum operator $bf J$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.
But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ bf J = bf X times bf P$$
Let's see that this is satisfied for $bf J_3$, i.e. $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
(other components will be analogous).
Let us consider an rotation acting on a state $|psirangle = int psi(x) |xrangle dx$.
$$ exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = int psi(x)|R(phi)xrangle dx = int psi(R^-1(phi)x) |xrangle$$
That means that
$$ bf J_3 |psirangle = ihbar left.fracddphiright|_phi=0 exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = ihbar int left.fracddphiright|_phi=0 psi(R^-1(phi)x) |xrangle$$
We have $$psi(R^-1(phi)x) = psi(x_1 cosphi + x_2 sinphi,-x_1sinphi+x_2cosphi,x_3) $$
so
$$ left.fracddphiright|_phi=0 psi(R^-1(phi)x) = x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2$$
so
beginalign bf J_3|psirangle &= ihbar int Big(x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2Big) |xrangle = \
&= -bf X_2 int Big(-ihbarfracpartial psi(x)partial x_1Big) |xrangle dx + bf X_1 int Big(-ihbarfracpartial psi(x)partial x_2Big) |xrangle = \ &= -bf X_2 bf P_1 int psi(x) |xrangle dx + bf X_1 bf P_2 int psi(x) |xrangle = \ &= (bf X_1bf P_2 - bf X_2bf P_1)|psirangle endalign
so $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
If a particle has internal degrees of freedom, then $$ bf J = bf X times bf P + bf S$$
where $bf S$ is the spin operator acting on these internal degrees of freedom.
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
$endgroup$
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
add a comment |
$begingroup$
Let's just consider $e^iJ_zphi$ where $hbar=1$.
We have
$$R_z=beginbmatrixcos(phi) & sin(phi)&0\-sin(phi)&cos(theta) & 0 \0&0&1endbmatrix$$
$$J_z=frac1ifracdR_z(phi)dphi|_phi=0=beginbmatrix0 & -i&0\i& 0 & 0 \0&0&0endbmatrix$$
We can exapnd exponential as:
$$e^iJ_zphi=1+iJ_zphi-J_z^2phi^2/2!-iJ_z^3phi^3/3!+....$$
Or
$$e^iJ_zphi=beginbmatrix1 & 0&0\0 & 1 & 0 \0&0&1endbmatrix+
phi beginbmatrix0 & 1&0\-1 & 0 & 0 \0&0&0endbmatrix
+phi^2/2! beginbmatrix-1 & 0&0\0& -1 & 0 \0&0&0endbmatrix+ ...=R_z(phi)$$
We can do the same for other axes.
Check out Quantum Field Theory by Lewis H. Ryder/
Single-particle relativistic wave equation.
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
Consider a smooth function $psi$ in $L^2(mathbbR^3, d^3x)$ (more precisely $psi$ is assumed to belong to Schwartz' space) and define for a fixed $n in mathbbS^2$ and $phi in mathbbR$
$$psi_phi(x)= psi(R^-1_n(phi)x)$$
Since $R^-1_n(phi) = e^-phi n cdot S$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$:
$$(S_j)_rs= epsilon_jrs$$
we have
$$fracddphipsi_phi(x) = -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi_phi(x):.tag1$$
Now consider
$$psi'_phi(x)= left(e^fracihbarphi n cdot Jpsiright)(x):,$$
where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only,
$$J_j = sum_r,s=1^3epsilon_jrs X_r P_s = -ihbar sum_r,s=1^3epsilon_jrs X_r fracpartialpartial x_s $$
Computing the $phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $psi$ matters), we have
$$fracddphipsi'_phi(x) = fracihbarsum_j=1^3 n cdot left(J_jpsi'_phiright)(x)= -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi'_phi(x):.tag2$$
In summary, for $x$ fixed, $psi_phi(x)$ and $psi'_phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition
$$psi_0(x)=psi'_0 (x):.$$
The uniqueness theorem for the solutions of first-order differential equations implies that
$$psi_phi(x)=psi'_phi (x):.$$
In other words
$$left(e^-fracihbar phi n cdot Jpsiright)(x) = psileft(R_n^-1(phi)xright):.$$
(The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.)
The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|yrangle = delta(x-y)= delta_y(x)$, so that, using rotational invariance of the Dirac delta function,
$$left(e^-fracihbar phi n cdot Jdelta_yright)(x) = delta_yleft(R_n^-1(phi)xright)= deltaleft(R_n^-1(phi)x -yright)= deltaleft(x -R_n(phi)yright):.$$
Coming back to the abstract notation, the found identity reads
$$e^-fracihbar phi n cdot J|yrangle = |R_n(phi)yrangle:.$$
(A completely rigorous proof can be found in this book of mine.)
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider a smooth function $psi$ in $L^2(mathbbR^3, d^3x)$ (more precisely $psi$ is assumed to belong to Schwartz' space) and define for a fixed $n in mathbbS^2$ and $phi in mathbbR$
$$psi_phi(x)= psi(R^-1_n(phi)x)$$
Since $R^-1_n(phi) = e^-phi n cdot S$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$:
$$(S_j)_rs= epsilon_jrs$$
we have
$$fracddphipsi_phi(x) = -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi_phi(x):.tag1$$
Now consider
$$psi'_phi(x)= left(e^fracihbarphi n cdot Jpsiright)(x):,$$
where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only,
$$J_j = sum_r,s=1^3epsilon_jrs X_r P_s = -ihbar sum_r,s=1^3epsilon_jrs X_r fracpartialpartial x_s $$
Computing the $phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $psi$ matters), we have
$$fracddphipsi'_phi(x) = fracihbarsum_j=1^3 n cdot left(J_jpsi'_phiright)(x)= -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi'_phi(x):.tag2$$
In summary, for $x$ fixed, $psi_phi(x)$ and $psi'_phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition
$$psi_0(x)=psi'_0 (x):.$$
The uniqueness theorem for the solutions of first-order differential equations implies that
$$psi_phi(x)=psi'_phi (x):.$$
In other words
$$left(e^-fracihbar phi n cdot Jpsiright)(x) = psileft(R_n^-1(phi)xright):.$$
(The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.)
The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|yrangle = delta(x-y)= delta_y(x)$, so that, using rotational invariance of the Dirac delta function,
$$left(e^-fracihbar phi n cdot Jdelta_yright)(x) = delta_yleft(R_n^-1(phi)xright)= deltaleft(R_n^-1(phi)x -yright)= deltaleft(x -R_n(phi)yright):.$$
Coming back to the abstract notation, the found identity reads
$$e^-fracihbar phi n cdot J|yrangle = |R_n(phi)yrangle:.$$
(A completely rigorous proof can be found in this book of mine.)
answered Jul 14 at 10:42
Valter MorettiValter Moretti
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Consider a smooth function $psi$ in $L^2(mathbbR^3, d^3x)$ (more precisely $psi$ is assumed to belong to Schwartz' space) and define for a fixed $n in mathbbS^2$ and $phi in mathbbR$
$$psi_phi(x)= psi(R^-1_n(phi)x)$$
Since $R^-1_n(phi) = e^-phi n cdot S$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$:
$$(S_j)_rs= epsilon_jrs$$
we have
$$fracddphipsi_phi(x) = -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi_phi(x):.tag1$$
Now consider
$$psi'_phi(x)= left(e^fracihbarphi n cdot Jpsiright)(x):,$$
where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only,
$$J_j = sum_r,s=1^3epsilon_jrs X_r P_s = -ihbar sum_r,s=1^3epsilon_jrs X_r fracpartialpartial x_s $$
Computing the $phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $psi$ matters), we have
$$fracddphipsi'_phi(x) = fracihbarsum_j=1^3 n cdot left(J_jpsi'_phiright)(x)= -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi'_phi(x):.tag2$$
In summary, for $x$ fixed, $psi_phi(x)$ and $psi'_phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition
$$psi_0(x)=psi'_0 (x):.$$
The uniqueness theorem for the solutions of first-order differential equations implies that
$$psi_phi(x)=psi'_phi (x):.$$
In other words
$$left(e^-fracihbar phi n cdot Jpsiright)(x) = psileft(R_n^-1(phi)xright):.$$
(The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.)
The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|yrangle = delta(x-y)= delta_y(x)$, so that, using rotational invariance of the Dirac delta function,
$$left(e^-fracihbar phi n cdot Jdelta_yright)(x) = delta_yleft(R_n^-1(phi)xright)= deltaleft(R_n^-1(phi)x -yright)= deltaleft(x -R_n(phi)yright):.$$
Coming back to the abstract notation, the found identity reads
$$e^-fracihbar phi n cdot J|yrangle = |R_n(phi)yrangle:.$$
(A completely rigorous proof can be found in this book of mine.)
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
$endgroup$
Consider a smooth function $psi$ in $L^2(mathbbR^3, d^3x)$ (more precisely $psi$ is assumed to belong to Schwartz' space) and define for a fixed $n in mathbbS^2$ and $phi in mathbbR$
$$psi_phi(x)= psi(R^-1_n(phi)x)$$
Since $R^-1_n(phi) = e^-phi n cdot S$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$:
$$(S_j)_rs= epsilon_jrs$$
we have
$$fracddphipsi_phi(x) = -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi_phi(x):.tag1$$
Now consider
$$psi'_phi(x)= left(e^fracihbarphi n cdot Jpsiright)(x):,$$
where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only,
$$J_j = sum_r,s=1^3epsilon_jrs X_r P_s = -ihbar sum_r,s=1^3epsilon_jrs X_r fracpartialpartial x_s $$
Computing the $phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $psi$ matters), we have
$$fracddphipsi'_phi(x) = fracihbarsum_j=1^3 n cdot left(J_jpsi'_phiright)(x)= -sum_j,r,s=1^3 n_j epsilon_jrsx_rfracpartial partial x_spsi'_phi(x):.tag2$$
In summary, for $x$ fixed, $psi_phi(x)$ and $psi'_phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition
$$psi_0(x)=psi'_0 (x):.$$
The uniqueness theorem for the solutions of first-order differential equations implies that
$$psi_phi(x)=psi'_phi (x):.$$
In other words
$$left(e^-fracihbar phi n cdot Jpsiright)(x) = psileft(R_n^-1(phi)xright):.$$
(The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.)
The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|yrangle = delta(x-y)= delta_y(x)$, so that, using rotational invariance of the Dirac delta function,
$$left(e^-fracihbar phi n cdot Jdelta_yright)(x) = delta_yleft(R_n^-1(phi)xright)= deltaleft(R_n^-1(phi)x -yright)= deltaleft(x -R_n(phi)yright):.$$
Coming back to the abstract notation, the found identity reads
$$e^-fracihbar phi n cdot J|yrangle = |R_n(phi)yrangle:.$$
(A completely rigorous proof can be found in this book of mine.)
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
edited Jul 14 at 11:15
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
answered Jul 14 at 10:42
Valter MorettiValter Moretti
38.1k4 gold badges73 silver badges145 bronze badges
38.1k4 gold badges73 silver badges145 bronze badges
add a comment |
add a comment |
$begingroup$
The most broad definition of the angular momentum operator $bf J$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.
But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ bf J = bf X times bf P$$
Let's see that this is satisfied for $bf J_3$, i.e. $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
(other components will be analogous).
Let us consider an rotation acting on a state $|psirangle = int psi(x) |xrangle dx$.
$$ exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = int psi(x)|R(phi)xrangle dx = int psi(R^-1(phi)x) |xrangle$$
That means that
$$ bf J_3 |psirangle = ihbar left.fracddphiright|_phi=0 exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = ihbar int left.fracddphiright|_phi=0 psi(R^-1(phi)x) |xrangle$$
We have $$psi(R^-1(phi)x) = psi(x_1 cosphi + x_2 sinphi,-x_1sinphi+x_2cosphi,x_3) $$
so
$$ left.fracddphiright|_phi=0 psi(R^-1(phi)x) = x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2$$
so
beginalign bf J_3|psirangle &= ihbar int Big(x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2Big) |xrangle = \
&= -bf X_2 int Big(-ihbarfracpartial psi(x)partial x_1Big) |xrangle dx + bf X_1 int Big(-ihbarfracpartial psi(x)partial x_2Big) |xrangle = \ &= -bf X_2 bf P_1 int psi(x) |xrangle dx + bf X_1 bf P_2 int psi(x) |xrangle = \ &= (bf X_1bf P_2 - bf X_2bf P_1)|psirangle endalign
so $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
If a particle has internal degrees of freedom, then $$ bf J = bf X times bf P + bf S$$
where $bf S$ is the spin operator acting on these internal degrees of freedom.
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
$endgroup$
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
add a comment |
$begingroup$
The most broad definition of the angular momentum operator $bf J$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.
But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ bf J = bf X times bf P$$
Let's see that this is satisfied for $bf J_3$, i.e. $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
(other components will be analogous).
Let us consider an rotation acting on a state $|psirangle = int psi(x) |xrangle dx$.
$$ exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = int psi(x)|R(phi)xrangle dx = int psi(R^-1(phi)x) |xrangle$$
That means that
$$ bf J_3 |psirangle = ihbar left.fracddphiright|_phi=0 exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = ihbar int left.fracddphiright|_phi=0 psi(R^-1(phi)x) |xrangle$$
We have $$psi(R^-1(phi)x) = psi(x_1 cosphi + x_2 sinphi,-x_1sinphi+x_2cosphi,x_3) $$
so
$$ left.fracddphiright|_phi=0 psi(R^-1(phi)x) = x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2$$
so
beginalign bf J_3|psirangle &= ihbar int Big(x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2Big) |xrangle = \
&= -bf X_2 int Big(-ihbarfracpartial psi(x)partial x_1Big) |xrangle dx + bf X_1 int Big(-ihbarfracpartial psi(x)partial x_2Big) |xrangle = \ &= -bf X_2 bf P_1 int psi(x) |xrangle dx + bf X_1 bf P_2 int psi(x) |xrangle = \ &= (bf X_1bf P_2 - bf X_2bf P_1)|psirangle endalign
so $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
If a particle has internal degrees of freedom, then $$ bf J = bf X times bf P + bf S$$
where $bf S$ is the spin operator acting on these internal degrees of freedom.
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
$endgroup$
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
add a comment |
$begingroup$
The most broad definition of the angular momentum operator $bf J$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.
But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ bf J = bf X times bf P$$
Let's see that this is satisfied for $bf J_3$, i.e. $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
(other components will be analogous).
Let us consider an rotation acting on a state $|psirangle = int psi(x) |xrangle dx$.
$$ exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = int psi(x)|R(phi)xrangle dx = int psi(R^-1(phi)x) |xrangle$$
That means that
$$ bf J_3 |psirangle = ihbar left.fracddphiright|_phi=0 exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = ihbar int left.fracddphiright|_phi=0 psi(R^-1(phi)x) |xrangle$$
We have $$psi(R^-1(phi)x) = psi(x_1 cosphi + x_2 sinphi,-x_1sinphi+x_2cosphi,x_3) $$
so
$$ left.fracddphiright|_phi=0 psi(R^-1(phi)x) = x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2$$
so
beginalign bf J_3|psirangle &= ihbar int Big(x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2Big) |xrangle = \
&= -bf X_2 int Big(-ihbarfracpartial psi(x)partial x_1Big) |xrangle dx + bf X_1 int Big(-ihbarfracpartial psi(x)partial x_2Big) |xrangle = \ &= -bf X_2 bf P_1 int psi(x) |xrangle dx + bf X_1 bf P_2 int psi(x) |xrangle = \ &= (bf X_1bf P_2 - bf X_2bf P_1)|psirangle endalign
so $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
If a particle has internal degrees of freedom, then $$ bf J = bf X times bf P + bf S$$
where $bf S$ is the spin operator acting on these internal degrees of freedom.
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
$endgroup$
The most broad definition of the angular momentum operator $bf J$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.
But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ bf J = bf X times bf P$$
Let's see that this is satisfied for $bf J_3$, i.e. $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
(other components will be analogous).
Let us consider an rotation acting on a state $|psirangle = int psi(x) |xrangle dx$.
$$ exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = int psi(x)|R(phi)xrangle dx = int psi(R^-1(phi)x) |xrangle$$
That means that
$$ bf J_3 |psirangle = ihbar left.fracddphiright|_phi=0 exp(- fracihbarphi, bf J_3) int psi(x)|xrangle dx = ihbar int left.fracddphiright|_phi=0 psi(R^-1(phi)x) |xrangle$$
We have $$psi(R^-1(phi)x) = psi(x_1 cosphi + x_2 sinphi,-x_1sinphi+x_2cosphi,x_3) $$
so
$$ left.fracddphiright|_phi=0 psi(R^-1(phi)x) = x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2$$
so
beginalign bf J_3|psirangle &= ihbar int Big(x_2 fracpartial psi(x)partial x_1 - x_1 fracpartial psi(x)partial x_2Big) |xrangle = \
&= -bf X_2 int Big(-ihbarfracpartial psi(x)partial x_1Big) |xrangle dx + bf X_1 int Big(-ihbarfracpartial psi(x)partial x_2Big) |xrangle = \ &= -bf X_2 bf P_1 int psi(x) |xrangle dx + bf X_1 bf P_2 int psi(x) |xrangle = \ &= (bf X_1bf P_2 - bf X_2bf P_1)|psirangle endalign
so $$ bf J_3 = bf X_1bf P_2 - bf X_2bf P_1$$
If a particle has internal degrees of freedom, then $$ bf J = bf X times bf P + bf S$$
where $bf S$ is the spin operator acting on these internal degrees of freedom.
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
answered Jul 14 at 11:14
Adam LatosińskiAdam Latosiński
4204 bronze badges
4204 bronze badges
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
add a comment |
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
$begingroup$
Thank you very much. Together with @Valter Morettis proof this proves the equality.
$endgroup$
– TheoreticalMinimum
Jul 14 at 11:22
add a comment |
$begingroup$
Let's just consider $e^iJ_zphi$ where $hbar=1$.
We have
$$R_z=beginbmatrixcos(phi) & sin(phi)&0\-sin(phi)&cos(theta) & 0 \0&0&1endbmatrix$$
$$J_z=frac1ifracdR_z(phi)dphi|_phi=0=beginbmatrix0 & -i&0\i& 0 & 0 \0&0&0endbmatrix$$
We can exapnd exponential as:
$$e^iJ_zphi=1+iJ_zphi-J_z^2phi^2/2!-iJ_z^3phi^3/3!+....$$
Or
$$e^iJ_zphi=beginbmatrix1 & 0&0\0 & 1 & 0 \0&0&1endbmatrix+
phi beginbmatrix0 & 1&0\-1 & 0 & 0 \0&0&0endbmatrix
+phi^2/2! beginbmatrix-1 & 0&0\0& -1 & 0 \0&0&0endbmatrix+ ...=R_z(phi)$$
We can do the same for other axes.
Check out Quantum Field Theory by Lewis H. Ryder/
Single-particle relativistic wave equation.
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's just consider $e^iJ_zphi$ where $hbar=1$.
We have
$$R_z=beginbmatrixcos(phi) & sin(phi)&0\-sin(phi)&cos(theta) & 0 \0&0&1endbmatrix$$
$$J_z=frac1ifracdR_z(phi)dphi|_phi=0=beginbmatrix0 & -i&0\i& 0 & 0 \0&0&0endbmatrix$$
We can exapnd exponential as:
$$e^iJ_zphi=1+iJ_zphi-J_z^2phi^2/2!-iJ_z^3phi^3/3!+....$$
Or
$$e^iJ_zphi=beginbmatrix1 & 0&0\0 & 1 & 0 \0&0&1endbmatrix+
phi beginbmatrix0 & 1&0\-1 & 0 & 0 \0&0&0endbmatrix
+phi^2/2! beginbmatrix-1 & 0&0\0& -1 & 0 \0&0&0endbmatrix+ ...=R_z(phi)$$
We can do the same for other axes.
Check out Quantum Field Theory by Lewis H. Ryder/
Single-particle relativistic wave equation.
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's just consider $e^iJ_zphi$ where $hbar=1$.
We have
$$R_z=beginbmatrixcos(phi) & sin(phi)&0\-sin(phi)&cos(theta) & 0 \0&0&1endbmatrix$$
$$J_z=frac1ifracdR_z(phi)dphi|_phi=0=beginbmatrix0 & -i&0\i& 0 & 0 \0&0&0endbmatrix$$
We can exapnd exponential as:
$$e^iJ_zphi=1+iJ_zphi-J_z^2phi^2/2!-iJ_z^3phi^3/3!+....$$
Or
$$e^iJ_zphi=beginbmatrix1 & 0&0\0 & 1 & 0 \0&0&1endbmatrix+
phi beginbmatrix0 & 1&0\-1 & 0 & 0 \0&0&0endbmatrix
+phi^2/2! beginbmatrix-1 & 0&0\0& -1 & 0 \0&0&0endbmatrix+ ...=R_z(phi)$$
We can do the same for other axes.
Check out Quantum Field Theory by Lewis H. Ryder/
Single-particle relativistic wave equation.
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
$endgroup$
Let's just consider $e^iJ_zphi$ where $hbar=1$.
We have
$$R_z=beginbmatrixcos(phi) & sin(phi)&0\-sin(phi)&cos(theta) & 0 \0&0&1endbmatrix$$
$$J_z=frac1ifracdR_z(phi)dphi|_phi=0=beginbmatrix0 & -i&0\i& 0 & 0 \0&0&0endbmatrix$$
We can exapnd exponential as:
$$e^iJ_zphi=1+iJ_zphi-J_z^2phi^2/2!-iJ_z^3phi^3/3!+....$$
Or
$$e^iJ_zphi=beginbmatrix1 & 0&0\0 & 1 & 0 \0&0&1endbmatrix+
phi beginbmatrix0 & 1&0\-1 & 0 & 0 \0&0&0endbmatrix
+phi^2/2! beginbmatrix-1 & 0&0\0& -1 & 0 \0&0&0endbmatrix+ ...=R_z(phi)$$
We can do the same for other axes.
Check out Quantum Field Theory by Lewis H. Ryder/
Single-particle relativistic wave equation.
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
edited Jul 14 at 10:58
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
answered Jul 14 at 10:53
ParadoxyParadoxy
6873 silver badges13 bronze badges
6873 silver badges13 bronze badges
add a comment |
add a comment |
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1
$begingroup$
is all of $hboldsymbolphi J$ in the denominator?
$endgroup$
– Luyw
Jul 14 at 7:49
$begingroup$
Shouldn't it be $hbar$ instead of $h$ ?
$endgroup$
– Thomas Fritsch
Jul 14 at 8:16
$begingroup$
I tried to address the misunderstandings.
$endgroup$
– TheoreticalMinimum
Jul 14 at 9:09
3
$begingroup$
I believe that the most broad definition of the angular momentum operator $bf J$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it.
$endgroup$
– Adam Latosiński
Jul 14 at 9:42
1
$begingroup$
@AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question.
$endgroup$
– Valter Moretti
Jul 14 at 10:05