When is separating the total wavefunction into a space part and a spin part possible?












5












$begingroup$


The total wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s}$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$. In my notation, $s=1/2, m_s=pm 1/2$.



Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.



Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.



Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.






quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    The total wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s}$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$. In my notation, $s=1/2, m_s=pm 1/2$.



    Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.



    Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.



    Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.






    quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      The total wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s}$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$. In my notation, $s=1/2, m_s=pm 1/2$.



      Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.



      Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.



      Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.






      quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles







      share|cite|improve this question











      $endgroup$




      The total wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s}$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$. In my notation, $s=1/2, m_s=pm 1/2$.



      Question 1 Is the above statement true? I am asking about any wavefunction here. Not only about energy eigenfunctions.



      Now imagine a system of two electrons. Even without any knowledge about the Hamiltonian of the system, the overall wavefunction $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ is antisymmetric. I think (I have this impression) under this general conditions, it is not possible to decompose $psi(vec{r}_1,vec{r}_2;s_1,s_2)$ into a product of a space part and spin part. However, if the Hamiltonian is spin-independent, only then can we do such a decomposition into space part and spin part.



      Question 2 Can someone properly argue that how this is so? Please mention about any wavefunction of the system and about energy eigenfunctions.






      quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles




      quantum-mechanics wavefunction quantum-spin pauli-exclusion-principle identical-particles






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago







      mithusengupta123

















      asked 2 days ago









      mithusengupta123mithusengupta123

      1,32311539




      1,32311539






















          1 Answer
          1






          active

          oldest

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          11












          $begingroup$

          Your claim




          [any arbitrary] wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s} tag 1$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$




          is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
          $$
          psi = frac{1}{sqrt{2}}bigg[phi_1zeta_1+phi_2zeta_2 bigg].
          $$



          Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.



          The result you want is the following:




          If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrm{space}otimes mathbb I+ mathbb I otimes H_mathrm{spin},$$ with $H_mathrm{space}otimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.




          To build that eigenbasis, simply diagonalize $H_mathrm{space}$ and $H_mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
            $endgroup$
            – Emilio Pisanty
            2 days ago












          • $begingroup$
            Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
            $endgroup$
            – user1936752
            2 days ago










          • $begingroup$
            @user1936752 Yes, this is redundant, but I don't think it hurts.
            $endgroup$
            – Emilio Pisanty
            2 days ago











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          11












          $begingroup$

          Your claim




          [any arbitrary] wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s} tag 1$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$




          is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
          $$
          psi = frac{1}{sqrt{2}}bigg[phi_1zeta_1+phi_2zeta_2 bigg].
          $$



          Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.



          The result you want is the following:




          If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrm{space}otimes mathbb I+ mathbb I otimes H_mathrm{spin},$$ with $H_mathrm{space}otimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.




          To build that eigenbasis, simply diagonalize $H_mathrm{space}$ and $H_mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
            $endgroup$
            – Emilio Pisanty
            2 days ago












          • $begingroup$
            Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
            $endgroup$
            – user1936752
            2 days ago










          • $begingroup$
            @user1936752 Yes, this is redundant, but I don't think it hurts.
            $endgroup$
            – Emilio Pisanty
            2 days ago
















          11












          $begingroup$

          Your claim




          [any arbitrary] wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s} tag 1$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$




          is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
          $$
          psi = frac{1}{sqrt{2}}bigg[phi_1zeta_1+phi_2zeta_2 bigg].
          $$



          Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.



          The result you want is the following:




          If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrm{space}otimes mathbb I+ mathbb I otimes H_mathrm{spin},$$ with $H_mathrm{space}otimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.




          To build that eigenbasis, simply diagonalize $H_mathrm{space}$ and $H_mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
            $endgroup$
            – Emilio Pisanty
            2 days ago












          • $begingroup$
            Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
            $endgroup$
            – user1936752
            2 days ago










          • $begingroup$
            @user1936752 Yes, this is redundant, but I don't think it hurts.
            $endgroup$
            – Emilio Pisanty
            2 days ago














          11












          11








          11





          $begingroup$

          Your claim




          [any arbitrary] wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s} tag 1$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$




          is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
          $$
          psi = frac{1}{sqrt{2}}bigg[phi_1zeta_1+phi_2zeta_2 bigg].
          $$



          Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.



          The result you want is the following:




          If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrm{space}otimes mathbb I+ mathbb I otimes H_mathrm{spin},$$ with $H_mathrm{space}otimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.




          To build that eigenbasis, simply diagonalize $H_mathrm{space}$ and $H_mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)






          share|cite|improve this answer











          $endgroup$



          Your claim




          [any arbitrary] wavefunction of an electron $psi(vec{r},s)$ can always be written as $$psi(vec{r},s)=phi(vec{r})zeta_{s,m_s} tag 1$$ where $phi(vec{r})$ is the space part and $zeta_{s,m_s}$ is the spin part of the total wavefunction $psi(vec{r},s)$




          is false. It is perfectly possible to produce wavefunctions which cannot be written in that separable form - for a simple example, just take two orthogonal spatial wavefunctions, $phi_1$ and $phi_2$, and two orthogonal spin states, $zeta_1$ and $zeta_2$, and define
          $$
          psi = frac{1}{sqrt{2}}bigg[phi_1zeta_1+phi_2zeta_2 bigg].
          $$



          Moreover, to be clear: the hamiltonian of a system has absolutely no effect on the allowed wavefunctions for that system. The only thing that depends on the hamiltonian is the energy eigenstates.



          The result you want is the following:




          If the hamiltonian is separable into spatial and spin components as $$ H = H_mathrm{space}otimes mathbb I+ mathbb I otimes H_mathrm{spin},$$ with $H_mathrm{space}otimes mathbb I$ commuting with all spin operators and $mathbb I otimes H_mathrm{spin}$ commuting with all space operators, then there exists an eigenbasis for $H$ of the separable form $(1)$.




          To build that eigenbasis, simply diagonalize $H_mathrm{space}$ and $H_mathrm{spin}$ independently, and form tensor products of their eigenstates. (Note also that the quantifiers here are crucial, particularly the "If" in the hypotheses and the "there exists" in the results.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Emilio PisantyEmilio Pisanty

          86.2k23213433




          86.2k23213433








          • 1




            $begingroup$
            @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
            $endgroup$
            – Emilio Pisanty
            2 days ago












          • $begingroup$
            Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
            $endgroup$
            – user1936752
            2 days ago










          • $begingroup$
            @user1936752 Yes, this is redundant, but I don't think it hurts.
            $endgroup$
            – Emilio Pisanty
            2 days ago














          • 1




            $begingroup$
            @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
            $endgroup$
            – Emilio Pisanty
            2 days ago












          • $begingroup$
            Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
            $endgroup$
            – Emilio Pisanty
            2 days ago










          • $begingroup$
            Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
            $endgroup$
            – user1936752
            2 days ago










          • $begingroup$
            @user1936752 Yes, this is redundant, but I don't think it hurts.
            $endgroup$
            – Emilio Pisanty
            2 days ago








          1




          1




          $begingroup$
          @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
          $endgroup$
          – Emilio Pisanty
          2 days ago




          $begingroup$
          @SRS The claim is specifically that there exists a separable eigenbasis. There is no claim that all eigenbases for such a hamiltonian are separable, because that claim is false. Please read more carefully.
          $endgroup$
          – Emilio Pisanty
          2 days ago












          $begingroup$
          @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
          $endgroup$
          – Emilio Pisanty
          2 days ago






          $begingroup$
          @SRS "H atom problem" is undefined. (I.e.: are you including fine structure and spin-orbit coupling?) If you're only talking about the Keplerian hamiltonian (i.e. kinetic energy plus electrostatic potential energy, with a frozen proton) then yes, a separable eigenbasis exists; there the $zeta_j$ are arbitrary (as there is no spin hamiltonian). If you're including spin-orbit coupling, then the hamiltonian does not satisfy the hypotheses I laid out, and the result is false.
          $endgroup$
          – Emilio Pisanty
          2 days ago














          $begingroup$
          Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
          $endgroup$
          – Emilio Pisanty
          2 days ago




          $begingroup$
          Comments are not for back-and-forth - particularly about another user's question. If you have further queries, take them to chat or ask separately.
          $endgroup$
          – Emilio Pisanty
          2 days ago












          $begingroup$
          Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
          $endgroup$
          – user1936752
          2 days ago




          $begingroup$
          Is the statement that all the $H_{space}otimes I$ must commute with all the $Iotimes H_{spin}$ not redundant? From the way you have written them, it seems like they must commute, no?
          $endgroup$
          – user1936752
          2 days ago












          $begingroup$
          @user1936752 Yes, this is redundant, but I don't think it hurts.
          $endgroup$
          – Emilio Pisanty
          2 days ago




          $begingroup$
          @user1936752 Yes, this is redundant, but I don't think it hurts.
          $endgroup$
          – Emilio Pisanty
          2 days ago


















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          迪纳利

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          迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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          南乌拉尔铁路局

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          南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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