Trjtdtk

Why is this a valid proof for the harmonic series?

I caught several of my students plagiarizing. Could it be my fault as a teacher?

How does NAND gate work? (Very basic question)

Python password manager

Binary Numbers Magic Trick

Airbnb - host wants to reduce rooms, can we get refund?

Junior developer struggles: how to communicate with management?

Is NMDA produced in the body?

Showing the sample mean is a sufficient statistics from an exponential distribution

Pressure to defend the relevance of one's area of mathematics

How to efficiently calculate prefix sum of frequencies of characters in a string?

How could a planet have most of its water in the atmosphere?

Is it cheaper to drop cargo than to land it?

CRT Oscilloscope - part of the plot is missing

Entropy as a function of temperature: is temperature well defined?

How to implement float hashing with approximate equality

How can I fairly adjudicate the effects of height differences on ranged attacks?

How to back up a running Linode server?

Is lying to get "gardening leave" fraud?

What does air vanishing on contact sound like?

Why is Thanos so tough at the beginning of "Avengers: Endgame"?

Meaning of "individuandum"

What is the most remote airport from the center of the city it supposedly serves?

Selecting a secure PIN for building access

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?

10

4

$begingroup$

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

share|cite|improve this question

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

$endgroup$

add a comment | 

10

4

$begingroup$

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

share|cite|improve this question

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

$endgroup$

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

share|cite|improve this question

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

$endgroup$

share|cite|improve this question

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

share|cite|improve this question

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

edited Apr 8 at 15:09

Qmechanic♦

108k122021255

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

asked Apr 8 at 13:01

SimoBartzSimoBartz

1068

2 Answers
2

active

oldest

votes

18

$begingroup$

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

share|cite|improve this answer

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer but it brings to another question, why do we call symmetry this condition?
  $endgroup$
  – SimoBartz
  Apr 8 at 13:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 8 at 16:58
  
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 8 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vectornaut What if they answered yes to any of those? What would you say?
  $endgroup$
  – opa
  Apr 9 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 10 at 12:54
  
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

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).

share|cite|improve this answer

answered Apr 8 at 13:18

LeviathanLeviathan

817

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f471292%2fsymmetry-in-quantum-mechanics%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

18

$begingroup$

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

share|cite|improve this answer

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer but it brings to another question, why do we call symmetry this condition?
  $endgroup$
  – SimoBartz
  Apr 8 at 13:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 8 at 16:58
  
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 8 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vectornaut What if they answered yes to any of those? What would you say?
  $endgroup$
  – opa
  Apr 9 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 10 at 12:54
  
  

  
  
  
  

 | 
show 2 more comments

18

$begingroup$

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

share|cite|improve this answer

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer but it brings to another question, why do we call symmetry this condition?
  $endgroup$
  – SimoBartz
  Apr 8 at 13:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 8 at 16:58
  
  

  
  
  
  


• 
  
  
  

  
  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
  Apr 8 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vectornaut What if they answered yes to any of those? What would you say?
  $endgroup$
  – opa
  Apr 9 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 10 at 12:54
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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

share|cite|improve this answer

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

answered Apr 8 at 13:12

Chiral AnomalyChiral Anomaly

14.6k22048

0

$begingroup$

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).

share|cite|improve this answer

answered Apr 8 at 13:18

LeviathanLeviathan

817

$endgroup$

add a comment | 

0

$begingroup$

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).

share|cite|improve this answer

answered Apr 8 at 13:18

LeviathanLeviathan

817

$endgroup$

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).

share|cite|improve this answer

answered Apr 8 at 13:18

LeviathanLeviathan

817

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:18

LeviathanLeviathan

817

answered Apr 8 at 13:18

LeviathanLeviathan

817

answered Apr 8 at 13:18

LeviathanLeviathan

817