Symmetry in quantum mechanics The 2019 Stack Overflow Developer Survey Results Are InSymmetry transformations on a quantum system; DefinitionsQuantum mechanics and Lorentz symmetryGenerators of a certain symmetry in Quantum MechanicsEquivalence of symmetry and commuting unitary operatorConcrete example that projective representation of symmetry group occurs in a quantum system except the case of spin half integer?Symmetry transformations on a quantum system; DefinitionsWhat is the definition of parity operator in quantum mechanics?Symmetry of Hamiltonian in harmonic oscillatorDifference between symmetry transformation and basis transformationSymmetries in quantum mechanicsWhat happens to the global $U(1)$ symmetry in alternative formulations of Quantum Mechanics?
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Symmetry in quantum mechanics
The 2019 Stack Overflow Developer Survey Results Are InSymmetry transformations on a quantum system; DefinitionsQuantum mechanics and Lorentz symmetryGenerators of a certain symmetry in Quantum MechanicsEquivalence of symmetry and commuting unitary operatorConcrete example that projective representation of symmetry group occurs in a quantum system except the case of spin half integer?Symmetry transformations on a quantum system; DefinitionsWhat is the definition of parity operator in quantum mechanics?Symmetry of Hamiltonian in harmonic oscillatorDifference between symmetry transformation and basis transformationSymmetries in quantum mechanicsWhat happens to the global $U(1)$ symmetry in alternative formulations of Quantum Mechanics?
$begingroup$
My professor told us that in quantum mechanics a transformation is a symmetry transformation if $$ UH(psi) = HU(psi) $$
Can you give me an easy explanation for this definition?
quantum-mechanics operators symmetry hamiltonian commutator
edited 2 days ago
Qmechanic♦
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
$endgroup$
add a comment |
$begingroup$
My professor told us that in quantum mechanics a transformation is a symmetry transformation if $$ UH(psi) = HU(psi) $$
Can you give me an easy explanation for this definition?
quantum-mechanics operators symmetry hamiltonian commutator
edited 2 days ago
Qmechanic♦
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
$endgroup$
add a comment |
$begingroup$
My professor told us that in quantum mechanics a transformation is a symmetry transformation if $$ UH(psi) = HU(psi) $$
Can you give me an easy explanation for this definition?
quantum-mechanics operators symmetry hamiltonian commutator
edited 2 days ago
Qmechanic♦
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
$endgroup$
My professor told us that in quantum mechanics a transformation is a symmetry transformation if $$ UH(psi) = HU(psi) $$
Can you give me an easy explanation for this definition?
quantum-mechanics operators symmetry hamiltonian commutator
quantum-mechanics operators symmetry hamiltonian commutator
edited 2 days ago
Qmechanic♦
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
edited 2 days ago
Qmechanic♦
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
edited 2 days ago
Qmechanic♦
107k122001241
edited 2 days ago
Qmechanic♦
107k122001241
edited 2 days ago
Qmechanic♦
107k122001241
107k122001241
asked 2 days ago
SimoBartzSimoBartz
1017
asked 2 days ago
SimoBartzSimoBartz
1017
asked 2 days ago
SimoBartzSimoBartz
1017
1017
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation
$$
ihbarfracddtpsi=Hpsi.
tag1
$$
We can multiply both sides of equation (1) by $U$ to get
$$
Uihbarfracddtpsi=UHpsi.
tag2
$$
If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
$$
ihbarfracddtUpsi=HUpsi.
tag3
$$
which says that if $psi$ solves equation (1), then so does $Upsi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
Symmetry transformations on a quantum system; Definitions
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
add a comment |
$begingroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator beginalign*
Uleft(t-t_0right)=e^-ileft(t-t_0right) H
endalign*
If $psileft(xi_1, dots, xi_N ; t_0right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
$left[Uleft(t-t_0right), Pright]=0$
then also
$$left(P Uleft(t-t_0right) psiright)left(xi_1, ldots, xi_N ; t_0right)=left(Uleft(t-t_0right) P psiright)left(xi_1, ldots, xi_N ; t_0right)$$
This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say:
beginalign*
[A, P]=0
endalign*
for all $Pin S_N$ (in permutation group of $N$ particles).
answered 2 days ago
LeviathanLeviathan
747
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation
$$
ihbarfracddtpsi=Hpsi.
tag1
$$
We can multiply both sides of equation (1) by $U$ to get
$$
Uihbarfracddtpsi=UHpsi.
tag2
$$
If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
$$
ihbarfracddtUpsi=HUpsi.
tag3
$$
which says that if $psi$ solves equation (1), then so does $Upsi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
Symmetry transformations on a quantum system; Definitions
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
add a comment |
$begingroup$
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation
$$
ihbarfracddtpsi=Hpsi.
tag1
$$
We can multiply both sides of equation (1) by $U$ to get
$$
Uihbarfracddtpsi=UHpsi.
tag2
$$
If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
$$
ihbarfracddtUpsi=HUpsi.
tag3
$$
which says that if $psi$ solves equation (1), then so does $Upsi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
Symmetry transformations on a quantum system; Definitions
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
add a comment |
$begingroup$
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation
$$
ihbarfracddtpsi=Hpsi.
tag1
$$
We can multiply both sides of equation (1) by $U$ to get
$$
Uihbarfracddtpsi=UHpsi.
tag2
$$
If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
$$
ihbarfracddtUpsi=HUpsi.
tag3
$$
which says that if $psi$ solves equation (1), then so does $Upsi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
Symmetry transformations on a quantum system; Definitions
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
$endgroup$
In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion.
In this case, the equation of motion is the Schrödinger equation
$$
ihbarfracddtpsi=Hpsi.
tag1
$$
We can multiply both sides of equation (1) by $U$ to get
$$
Uihbarfracddtpsi=UHpsi.
tag2
$$
If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
$$
ihbarfracddtUpsi=HUpsi.
tag3
$$
which says that if $psi$ solves equation (1), then so does $Upsi$, so $U$ is a symmetry.
For a more general definition of symmetry in QM, see
Symmetry transformations on a quantum system; Definitions
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
answered 2 days ago
Chiral AnomalyChiral Anomaly
13.5k21845
13.5k21845
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
add a comment |
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
3
3
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
$begingroup$
@SimoBartz That's a good question. In a more completely specified model, say with lots of local observables as in quantum field theory, we would require that a symmetry preserve things like the relationships between those observables in space and time. But in the present question, only the Hamiltonian is specified, so there is nothing else to preserve.
$endgroup$
– Chiral Anomaly
2 days ago
1
1
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@SimoBartz, what does the word "symmetry" mean to you? Have you encountered it in other contexts, such as classical mechanics or geometry?
$endgroup$
– Vectornaut
2 days ago
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
@Vectornaut What if they answered yes to any of those? What would you say?
$endgroup$
– opa
yesterday
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
$begingroup$
Actually I'have never seen this concept before, my professor told us that when you have a symmetry transformation the system is invariant respect to that transformation. I imagine it means that nothing changes except the point of view. But if I transform a solution in another one maybe the new solution is completely different
$endgroup$
– SimoBartz
16 hours ago
add a comment |
$begingroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator beginalign*
Uleft(t-t_0right)=e^-ileft(t-t_0right) H
endalign*
If $psileft(xi_1, dots, xi_N ; t_0right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
$left[Uleft(t-t_0right), Pright]=0$
then also
$$left(P Uleft(t-t_0right) psiright)left(xi_1, ldots, xi_N ; t_0right)=left(Uleft(t-t_0right) P psiright)left(xi_1, ldots, xi_N ; t_0right)$$
This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say:
beginalign*
[A, P]=0
endalign*
for all $Pin S_N$ (in permutation group of $N$ particles).
answered 2 days ago
LeviathanLeviathan
747
$endgroup$
add a comment |
$begingroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator beginalign*
Uleft(t-t_0right)=e^-ileft(t-t_0right) H
endalign*
If $psileft(xi_1, dots, xi_N ; t_0right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
$left[Uleft(t-t_0right), Pright]=0$
then also
$$left(P Uleft(t-t_0right) psiright)left(xi_1, ldots, xi_N ; t_0right)=left(Uleft(t-t_0right) P psiright)left(xi_1, ldots, xi_N ; t_0right)$$
This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say:
beginalign*
[A, P]=0
endalign*
for all $Pin S_N$ (in permutation group of $N$ particles).
answered 2 days ago
LeviathanLeviathan
747
$endgroup$
add a comment |
$begingroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator beginalign*
Uleft(t-t_0right)=e^-ileft(t-t_0right) H
endalign*
If $psileft(xi_1, dots, xi_N ; t_0right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
$left[Uleft(t-t_0right), Pright]=0$
then also
$$left(P Uleft(t-t_0right) psiright)left(xi_1, ldots, xi_N ; t_0right)=left(Uleft(t-t_0right) P psiright)left(xi_1, ldots, xi_N ; t_0right)$$
This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say:
beginalign*
[A, P]=0
endalign*
for all $Pin S_N$ (in permutation group of $N$ particles).
answered 2 days ago
LeviathanLeviathan
747
$endgroup$
What you have written there is nothing but the commutator. Consider for example the time evolution operator beginalign*
Uleft(t-t_0right)=e^-ileft(t-t_0right) H
endalign*
If $psileft(xi_1, dots, xi_N ; t_0right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
$left[Uleft(t-t_0right), Pright]=0$
then also
$$left(P Uleft(t-t_0right) psiright)left(xi_1, ldots, xi_N ; t_0right)=left(Uleft(t-t_0right) P psiright)left(xi_1, ldots, xi_N ; t_0right)$$
This means that the permuted time evolved wave function is the same as the time evolved permuted wave function.
Another example would be if you consider identical particles. An arbitrary observable $A$ should be the same under the permutation operator $P$ if one has identical particles. This is to say:
beginalign*
[A, P]=0
endalign*
for all $Pin S_N$ (in permutation group of $N$ particles).
answered 2 days ago
LeviathanLeviathan
747
answered 2 days ago
LeviathanLeviathan
747
answered 2 days ago
LeviathanLeviathan
747
answered 2 days ago
LeviathanLeviathan
747
747
add a comment |
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
