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 1 hour ago
Qmechanic♦
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
$endgroup$
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 1 hour ago
Qmechanic♦
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
$endgroup$
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 1 hour ago
Qmechanic♦
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
$endgroup$
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 1 hour ago
Qmechanic♦
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
edited 1 hour ago
Qmechanic♦
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
edited 1 hour ago
Qmechanic♦
107k121991239
edited 1 hour ago
Qmechanic♦
107k121991239
edited 1 hour ago
Qmechanic♦
107k121991239
107k121991239
asked 3 hours ago
SimoBartzSimoBartz
647
asked 3 hours ago
SimoBartzSimoBartz
647
asked 3 hours ago
SimoBartzSimoBartz
647
647
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 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
$endgroup$
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 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 hours 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 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
$endgroup$
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 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 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
$endgroup$
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 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 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
$endgroup$
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 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
answered 2 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
answered 2 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
answered 2 hours ago
Chiral AnomalyChiral Anomaly
13.3k21744
13.3k21744
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 hours ago
add a comment |
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 hours ago
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 hours ago
$begingroup$
This is a good answer but it brings to another question, why do we call symmetry this condition?
$endgroup$
– SimoBartz
2 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 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 2 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 2 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 2 hours ago
LeviathanLeviathan
747
answered 2 hours ago
LeviathanLeviathan
747
answered 2 hours ago
LeviathanLeviathan
747
answered 2 hours ago
LeviathanLeviathan
747
747
add a comment |
add a comment |
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