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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          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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          When can things happen in Etherscan, such as the picture below?

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          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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