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♦

      107k122001241




      107k122001241










      asked 2 days ago









      SimoBartzSimoBartz

      1017




      1017






















          2 Answers
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          active

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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
            15 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

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            active

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            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
              15 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
              15 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
              15 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
              15 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
            15 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
            15 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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                    GameSpot

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                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
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                    日野市

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                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
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                    Tu-95轟炸機

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                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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