Symmetry in quantum mechanicsSymmetry 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
Symmetry 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 11 hours ago
Qmechanic♦
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
$endgroup$
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 11 hours ago
Qmechanic♦
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
$endgroup$
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 11 hours ago
Qmechanic♦
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
$endgroup$
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 11 hours ago
Qmechanic♦
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
edited 11 hours ago
Qmechanic♦
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
edited 11 hours ago
Qmechanic♦
107k121991239
edited 11 hours ago
Qmechanic♦
107k121991239
edited 11 hours ago
Qmechanic♦
107k121991239
107k121991239
asked 13 hours ago
SimoBartzSimoBartz
897
asked 13 hours ago
SimoBartzSimoBartz
897
asked 13 hours ago
SimoBartzSimoBartz
897
897
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 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
$endgroup$
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
4 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 13 hours ago
LeviathanLeviathan
747
$endgroup$
add a comment |
Your Answer
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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 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
$endgroup$
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
4 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 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
$endgroup$
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
4 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 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
$endgroup$
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 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
answered 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
answered 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
answered 13 hours ago
Chiral AnomalyChiral Anomaly
13.4k21745
13.4k21745
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
4 hours ago
add a comment |
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
4 hours ago
2
2
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours ago
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
12 hours 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
9 hours 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
9 hours 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
4 hours 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
4 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 13 hours 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 13 hours 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 13 hours 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 13 hours ago
LeviathanLeviathan
747
answered 13 hours ago
LeviathanLeviathan
747
answered 13 hours ago
LeviathanLeviathan
747
answered 13 hours ago
LeviathanLeviathan
747
747
add a comment |
add a comment |
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