You have summarized 4 hours of class in 5 minutes. You are the best. I wish my quantum physics teachers would start explaining things this simple hahaha
How does one get the normalization constant N for Y(1,2)? |Y(1,2)|^2 = N^2 * (Y1(1)Y2(2)-Y1(2)Y2(1))^2 = N^2 * { (Y1(1)Y2(2))^2 - 2*Y1(1)Y2(2)*Y1(2)Y2(1) + (Y1(2)Y2(1))^2 }. The Terms in {} are Integrals. But how do they become 1 + 1? I think in general, I have troubles understanding when an integral of a product of wavefunctions is equal to 0 oder 1.
Hi Johanna. This is easiest to understand if we assume that we have a set of n orthonormal one-particle wavefunctions (it can still be done if that's not true, the algebra is just much more nasty). In that case, any time you have an integral over the coordinates of a particle, you get a `1` if i = j in the product Ys_i(n1)Y_j(n1) and a `0` if i =/= j. The result of the integration over the coordinates of every particle can be separated into the product of the integral over each individual particle (separation of variables), So Ys_i1(n1)Y_j1(n1)Ys_i2(n2)Y_j2(n2)...Ys_iN(nN)Y_jN(nN) is `1` if i1 = j1 and i2= j2 ... and iN = jN, and `0` if any of the indices don't match. What ends up happening is that in any N electron slater determinant your wavefunction has N! terms, and your probability density Ys(1,2,...,N)Y(1,2,...N) has (N!)^2 terms. In N! of them, all the indices are equal and you get a `1`, and in (N!)^2 - N! of them at least one set of indices are mismatched and you get a `0`. Thus the integral psi_star * psi for any Slater determined of orthonormal orbitals integrates to N! and normalizes with a pre-factor of 1/sqrt(N!).
First of all thanks a lot for such great video!!! However I believe you have committed a verbal mistake when you refer to columns and rows in the slater determinant....you mention at 2:15 that each row has different electrons while each column in fact has different electron. Please correct me if I am wrong?
Hi Amin. After further review I believe it was said correctly. Row 1 represents electron 1, and column 1 represents orbital one. Note how the orbital subscripts change in each column, whereas the electron index changes in each row.
@@TMPChem I'm confused, as my lecturer used the rows to represent the orbital and the column to represent the electrons, just as Amin said. Is there any way that I can work around this?
@@ph7613 Im my lecture notes, the columns represent the electrons, while for example at Wikipedia the rows do, but at the end of the day, it doesn't matter because the determinant of a transposed matrix is the same as the one of the original matrix.
You have summarized 4 hours of class in 5 minutes. You are the best. I wish my quantum physics teachers would start explaining things this simple hahaha
Best RUclips channel!
Thank you so much for the videos!
Really good lecture!!
Thank you so, so much!
Good on ya mate
How does one get the normalization constant N for Y(1,2)? |Y(1,2)|^2 = N^2 * (Y1(1)Y2(2)-Y1(2)Y2(1))^2 = N^2 * { (Y1(1)Y2(2))^2 - 2*Y1(1)Y2(2)*Y1(2)Y2(1) + (Y1(2)Y2(1))^2 }. The Terms in {} are Integrals. But how do they become 1 + 1? I think in general, I have troubles understanding when an integral of a product of wavefunctions is equal to 0 oder 1.
Hi Johanna. This is easiest to understand if we assume that we have a set of n orthonormal one-particle wavefunctions (it can still be done if that's not true, the algebra is just much more nasty). In that case, any time you have an integral over the coordinates of a particle, you get a `1` if i = j in the product Ys_i(n1)Y_j(n1) and a `0` if i =/= j. The result of the integration over the coordinates of every particle can be separated into the product of the integral over each individual particle (separation of variables), So Ys_i1(n1)Y_j1(n1)Ys_i2(n2)Y_j2(n2)...Ys_iN(nN)Y_jN(nN) is `1` if i1 = j1 and i2= j2 ... and iN = jN, and `0` if any of the indices don't match. What ends up happening is that in any N electron slater determinant your wavefunction has N! terms, and your probability density Ys(1,2,...,N)Y(1,2,...N) has (N!)^2 terms. In N! of them, all the indices are equal and you get a `1`, and in (N!)^2 - N! of them at least one set of indices are mismatched and you get a `0`. Thus the integral psi_star * psi for any Slater determined of orthonormal orbitals integrates to N! and normalizes with a pre-factor of 1/sqrt(N!).
Thank you!
What if we have 5 states and 2 electrons ,then how many 2-particle states are possible ? pls reply 🙏
First of all thanks a lot for such great video!!! However I believe you have committed a verbal mistake when you refer to columns and rows in the slater determinant....you mention at 2:15 that each row has different electrons while each column in fact has different electron. Please correct me if I am wrong?
Hi Amin. After further review I believe it was said correctly. Row 1 represents electron 1, and column 1 represents orbital one. Note how the orbital subscripts change in each column, whereas the electron index changes in each row.
@@TMPChem I'm confused, as my lecturer used the rows to represent the orbital and the column to represent the electrons, just as Amin said. Is there any way that I can work around this?
@@ph7613 Im my lecture notes, the columns represent the electrons, while for example at Wikipedia the rows do, but at the end of the day, it doesn't matter because the determinant of a transposed matrix is the same as the one of the original matrix.
Thank you so much
Thanks
BRUH
Yup.
@@TMPChem Yeet.