Symmetry in quantum mechanics












9












$begingroup$


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











$endgroup$

















    9












    $begingroup$


    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











    $endgroup$















      9












      9








      9


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Qmechanic♦

      107k121991241




      107k121991241










      asked 2 days ago









      SimoBartzSimoBartz

      1017




      1017






















          2 Answers
          2






          active

          oldest

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          17












          $begingroup$

          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
          $$
          ihbarfrac{d}{dt}psi=Hpsi.
          tag{1}
          $$
          We can multiply both sides of equation (1) by $U$ to get
          $$
          Uihbarfrac{d}{dt}psi=UHpsi.
          tag{2}
          $$
          If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
          $$
          ihbarfrac{d}{dt}Upsi=HUpsi.
          tag{3}
          $$
          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









          $endgroup$









          • 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
            13 hours ago





















          0












          $begingroup$

          What you have written there is nothing but the commutator. Consider for example the time evolution operator begin{align*}
          Uleft(t-t_{0}right)=e^{-ileft(t-t_{0}right) H}
          end{align*}
          If $psileft(xi_{1}, dots, xi_{N} ; t_{0}right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
          $left[Uleft(t-t_{0}right), Pright]=0$
          then also
          $$left(P Uleft(t-t_{0}right) psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)=left(Uleft(t-t_{0}right) P psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)$$
          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:
          begin{align*}
          [A, P]=0
          end{align*}
          for all $Pin S_N$ (in permutation group of $N$ particles).






          share|cite|improve this answer









          $endgroup$














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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

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            votes






            active

            oldest

            votes









            17












            $begingroup$

            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
            $$
            ihbarfrac{d}{dt}psi=Hpsi.
            tag{1}
            $$
            We can multiply both sides of equation (1) by $U$ to get
            $$
            Uihbarfrac{d}{dt}psi=UHpsi.
            tag{2}
            $$
            If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
            $$
            ihbarfrac{d}{dt}Upsi=HUpsi.
            tag{3}
            $$
            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









            $endgroup$









            • 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
              13 hours ago


















            17












            $begingroup$

            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
            $$
            ihbarfrac{d}{dt}psi=Hpsi.
            tag{1}
            $$
            We can multiply both sides of equation (1) by $U$ to get
            $$
            Uihbarfrac{d}{dt}psi=UHpsi.
            tag{2}
            $$
            If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
            $$
            ihbarfrac{d}{dt}Upsi=HUpsi.
            tag{3}
            $$
            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









            $endgroup$









            • 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
              13 hours ago
















            17












            17








            17





            $begingroup$

            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
            $$
            ihbarfrac{d}{dt}psi=Hpsi.
            tag{1}
            $$
            We can multiply both sides of equation (1) by $U$ to get
            $$
            Uihbarfrac{d}{dt}psi=UHpsi.
            tag{2}
            $$
            If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
            $$
            ihbarfrac{d}{dt}Upsi=HUpsi.
            tag{3}
            $$
            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









            $endgroup$



            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
            $$
            ihbarfrac{d}{dt}psi=Hpsi.
            tag{1}
            $$
            We can multiply both sides of equation (1) by $U$ to get
            $$
            Uihbarfrac{d}{dt}psi=UHpsi.
            tag{2}
            $$
            If $UH=HU$ and $U$ is independent of time, then equation (2) may be rewritten as
            $$
            ihbarfrac{d}{dt}Upsi=HUpsi.
            tag{3}
            $$
            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












            share|cite|improve this answer



            share|cite|improve this answer










            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
              13 hours ago
















            • 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
              13 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
            13 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
            13 hours ago













            0












            $begingroup$

            What you have written there is nothing but the commutator. Consider for example the time evolution operator begin{align*}
            Uleft(t-t_{0}right)=e^{-ileft(t-t_{0}right) H}
            end{align*}
            If $psileft(xi_{1}, dots, xi_{N} ; t_{0}right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
            $left[Uleft(t-t_{0}right), Pright]=0$
            then also
            $$left(P Uleft(t-t_{0}right) psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)=left(Uleft(t-t_{0}right) P psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)$$
            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:
            begin{align*}
            [A, P]=0
            end{align*}
            for all $Pin S_N$ (in permutation group of $N$ particles).






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              What you have written there is nothing but the commutator. Consider for example the time evolution operator begin{align*}
              Uleft(t-t_{0}right)=e^{-ileft(t-t_{0}right) H}
              end{align*}
              If $psileft(xi_{1}, dots, xi_{N} ; t_{0}right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
              $left[Uleft(t-t_{0}right), Pright]=0$
              then also
              $$left(P Uleft(t-t_{0}right) psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)=left(Uleft(t-t_{0}right) P psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)$$
              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:
              begin{align*}
              [A, P]=0
              end{align*}
              for all $Pin S_N$ (in permutation group of $N$ particles).






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                What you have written there is nothing but the commutator. Consider for example the time evolution operator begin{align*}
                Uleft(t-t_{0}right)=e^{-ileft(t-t_{0}right) H}
                end{align*}
                If $psileft(xi_{1}, dots, xi_{N} ; t_{0}right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
                $left[Uleft(t-t_{0}right), Pright]=0$
                then also
                $$left(P Uleft(t-t_{0}right) psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)=left(Uleft(t-t_{0}right) P psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)$$
                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:
                begin{align*}
                [A, P]=0
                end{align*}
                for all $Pin S_N$ (in permutation group of $N$ particles).






                share|cite|improve this answer









                $endgroup$



                What you have written there is nothing but the commutator. Consider for example the time evolution operator begin{align*}
                Uleft(t-t_{0}right)=e^{-ileft(t-t_{0}right) H}
                end{align*}
                If $psileft(xi_{1}, dots, xi_{N} ; t_{0}right)$ is the wave function at time $t_0$ and $U(t−t0)$ is the time evolution operator that for all permutations $P$ satisfies
                $left[Uleft(t-t_{0}right), Pright]=0$
                then also
                $$left(P Uleft(t-t_{0}right) psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)=left(Uleft(t-t_{0}right) P psiright)left(xi_{1}, ldots, xi_{N} ; t_{0}right)$$
                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:
                begin{align*}
                [A, P]=0
                end{align*}
                for all $Pin S_N$ (in permutation group of $N$ particles).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                LeviathanLeviathan

                747




                747






























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                    Tanganjiko

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                    Lago Tanganjiko Situo centra Afriko Ĉefaj fontoj Kalambo, Lufuko, Malagarasi kaj Ruzizi Ĉefaj elfluoj Lukugo Setlejoj Bujumbura, Kalemie Datenoj Koordinatoj 6°  S , 30°  O -6.3827777777778 29.615 782 Koordinatoj: 6°  S , 30°  O DEC Supermara alteco 782 m Surfaca areo 32 893 km² f5 Maksimuma longo 673 km f6 Maksimuma larĝo 72 km f7 Akva volumeno 18.900 km³ dep1 f8 Maksimuma profundo 1470 m f10 Averaĝa profundo 570 m f11 v   •   d   •   r Mapo de la lago Tanganjiko estas lago en orienta Afriko, inter Burundo, Kongo Kinŝasa, Tanzanio kaj Zambio. Ĝi apartenas al la Granda Rifto kaj estas unu el la Afrikaj Grandaj Lagoj. Tanzanio (46%) kaj DRK (40%) havas la majoritaton de la lago. Ĝi havas longon de 673 km, areon de ĉ. 32.900 km² kaj maksimuman profundon taksatan je 1470 metroj; pro tiu ĉi granda profundo ankaŭ ĝia volumeno estas sufiĉe granda, 18.900 km³. Oni konsi
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                    Liste der Baudenkmäler in Enneberg

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                    Die Liste der Baudenkmäler in Enneberg (ladinisch Mareo , italienisch Marebbe ) enthält die 36 als Baudenkmäler ausgewiesenen Objekte auf dem Gebiet der Gemeinde Enneberg in Südtirol. Basis ist das im Internet einsehbare offizielle Verzeichnis der Baudenkmäler in Südtirol. Dabei kann es sich beispielsweise um Sakralbauten, Wohnhäuser, Bauernhöfe und Adelsansitze handeln. Die Reihenfolge in dieser Liste orientiert sich an der Bezeichnung, alternativ ist sie auch nach der Adresse oder dem Datum der Unterschutzstellung sortierbar. Liste | Foto   Bezeichnung Standort Eintragung Beschreibung Metadaten ja Altes Gasthaus Janesch ID: 14567 46° 42′ 54″ N, 11° 52′ 42″ O 76882557800 0 8. Sep. 1986 (BLR-LAB 4974) Ländliches Wohnhaus/Bauernhof KG:  Welschellen Bauparzelle: 391 ja Asch (Prack) ID: 14540 46° 43′ 12″ N, 11° 54′ 35″ O 75496485300 20. Nov. 1950 (MD) Ansitz KG:  Enneberg Bauparzelle: 38 ja Biccia mit Stade
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