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 yesterday







      mithusengupta123

















      asked yesterday









      mithusengupta123mithusengupta123

      1,32311539




      1,32311539






















          1 Answer
          1






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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
            yesterday










          • $begingroup$
            @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
            yesterday












          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday










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











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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
            yesterday










          • $begingroup$
            @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
            yesterday












          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday










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
















          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
            yesterday










          • $begingroup$
            @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
            yesterday












          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday










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














          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 yesterday

























          answered yesterday









          Emilio PisantyEmilio Pisanty

          86.1k23213432




          86.1k23213432








          • 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
            yesterday










          • $begingroup$
            @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
            yesterday












          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday










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














          • 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
            yesterday










          • $begingroup$
            @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
            yesterday












          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday










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








          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
          yesterday




          $begingroup$
          @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
          yesterday












          $begingroup$
          @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
          yesterday






          $begingroup$
          @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
          yesterday














          $begingroup$
          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
          yesterday




          $begingroup$
          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
          yesterday












          $begingroup$
          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
          yesterday




          $begingroup$
          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
          yesterday












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




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


















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          Bunad

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          Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
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