# 物理代写|量子力学代写quantum mechanics代考|PHYSICS3544

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## 物理代写|量子力学代写quantum mechanics代考|Antisymmetric wave function and the Slater determinant

An alternating function has an antisymmetric property that by interchanging two variables it gives the value of the function multiplied by $-1$ (see Chapter thirteen).
$$f\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right)=-f\left(x_{2}, x_{1}, x_{3}, \ldots, x_{n}\right)$$

When the matrix elements are functions, $f_{\mathrm{i}}\left(\mathrm{x}{\mathrm{i}}\right.$ ), the second (or fourth or sixth and so on) term of the determinant of this matrix, $\mathrm{D}$, is an alternating function of its previous term of the determinant. The determinant of order $\mathrm{n}$ has $\mathrm{n}$ ! permutations, i.e., operations of changing two rows or two columns in the matrix that give the minus determinant of the orginal matrix. \begin{aligned} D{2 \times 2} &=\left|\begin{array}{ll} f_{1}\left(x_{1}\right) & f_{1}\left(x_{2}\right) \ f_{2}\left(x_{1}\right) & f_{2}\left(x_{2}\right) \end{array}\right|=f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right)-f_{1}\left(x_{2}\right) f_{2}\left(x_{1}\right) \ D_{3 \times 3} &=\left|\begin{array}{lll} f_{1}\left(x_{1}\right) & f_{1}\left(x_{2}\right) & f_{1}\left(x_{3}\right) \ f_{2}\left(x_{1}\right) & f_{2}\left(x_{2}\right) & f_{2}\left(x_{3}\right) \ f_{3}\left(x_{1}\right) & f_{3}\left(x_{2}\right) & f_{3}\left(x_{3}\right) \end{array}\right|=\left{\begin{array}{l} f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right) f_{3}\left(x_{3}\right)-f_{1}\left(x_{1}\right) f_{2}\left(x_{3}\right) f_{3}\left(x_{2}\right) \ f_{1}\left(x_{2}\right) f_{2}\left(x_{3}\right) f_{3}\left(x_{1}\right)-f_{1}\left(x_{2}\right) f_{2}\left(x_{1}\right) f_{3}\left(x_{3}\right) \ f_{1}\left(x_{3}\right) f_{2}\left(x_{1}\right) f_{3}\left(x_{2}\right)-f_{1}\left(x_{3}\right) f_{2}\left(x_{2}\right) f_{3}\left(x_{1}\right) \end{array}\right. \end{aligned}
If the matrix element, $f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}\right)$, is a spin-orbital, then the determinant is called Slater determinant, although it was firstly used by Heisenberg and Dirac (see Chapter thirteen). Alternating functions are important to construct the antisymmetric wave function.

## 物理代写|量子力学代写quantum mechanics代考|Properties of determinants and the Slater determinant

If two rows or two columns of a determinant, $\mathrm{D}$, are equal, then $\mathrm{D}=0$. Pauli exclusion principle states where two electrons are in the same orbital, then they cannot have the same spin-orbital. Then, in the Slater determinant, two rows or two spin-orbitals cannot be the same, i.e., $f_{1} \neq f_{2}$, otherwise, the Slater determinant (which represents the set of molecular orbital, MO, wave functions) is zero.

To sum up, the properties of the determinants for square matrices (including comments on the Slater determinants, i.e., the MO wave functions), we have:
(i) If any row of column of the matrix has only zero elements, its determinant is zero (there cannot exist one orbital with zero elements, even LUMO orbital has to have non zero elements, because otherwise the Slater determinant is zero).
(ii) The determinant of a matrix, $\operatorname{det}(\mathbf{A})$, gives the same determinant with minus sign, $-\operatorname{det}(\hat{A})$, after the operation $\left(\mathrm{E}{\mathrm{i}}\right) \leftrightarrow\left(\mathrm{E}{\mathrm{k}}\right)$, with $\mathrm{i} \neq \mathrm{k}$. This corresponds to the antisymmetric property of a fermion (previous section). That is: $\operatorname{det}(\boldsymbol{A})=$ $-\operatorname{det}(\hat{A})$, where $\hat{\mathrm{A}}:\left(\mathrm{E}{\mathrm{i}}\right) \leftrightarrow\left(\mathrm{E}{\mathrm{k}}\right)$
(iii) If a matrix has two rows or two columns with the same elements, then its determinant is zero.
(iv) If a matrix is obtained after the operation $\left(\lambda \mathrm{E}{\mathrm{i}}\right) \rightarrow\left(\mathrm{E}{\mathrm{i}}\right)$, then: $\operatorname{det}(\hat{A})=\lambda \operatorname{det}(A)$, where $\hat{\mathrm{A}}:\left(\lambda \mathrm{E}{\mathrm{i}}\right) \rightarrow\left(\mathrm{E}{\mathrm{i}}\right)$.
(v) If a matrix is obtained from the operation $\left(\mathrm{E}{\mathrm{i}}+\lambda \mathrm{E}{\mathrm{k}}\right) \rightarrow\left(\mathrm{E}{\mathrm{i}}\right)$, with $\mathrm{i} \neq \mathrm{k}$, then: $\operatorname{det}(\hat{A})=\operatorname{det}(\boldsymbol{A})$, where $\hat{\mathrm{A}}:\left(\mathrm{E}{\mathrm{i}}+\lambda \mathrm{E}{\mathrm{k}}\right) \rightarrow\left(\mathrm{E}{\mathrm{i}}\right)$. The Slater determinant is the same after changing two orbitals by $\left(\mathrm{E}{\mathrm{i}}+\lambda \mathrm{E}{\mathrm{k}}\right) \rightarrow\left(\mathrm{E}_{\mathrm{i}}\right)$.
(vi) If $\mathbf{A}$ and $\mathbf{B}$ are square matrices of the same order, then $\operatorname{det}(\boldsymbol{A B})=\operatorname{det}(\boldsymbol{A}) \operatorname{det}(\boldsymbol{B})$
(vii) The determinant of the transpose of a matrix is the same as the determinant of the original matrix, $\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)$. Then, the MO orbitals can be in rows or columns of the Slater determinant.

# 量子力学代考

## 物理代写|量子力学代写quantum mechanics代考|Antisymmetric wave function and the Slater determinant

$$f\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right)=-f\left(x_{2}, x_{1}, x_{3}, \ldots, x_{n}\right)$$

$$f_{1}\left(x_{1}\right) \quad f_{1}\left(x_{2}\right) f_{2}\left(x_{1}\right) \quad f_{2}\left(x_{2}\right)$$
}\left(x_{1}〉right) \
$D_{-}{3 \backslash$ times 3$}$ \& $=\backslash$ left $\mid$
$f_{1}\left(x_{1}\right) \quad f_{1}\left(x_{2}\right) \quad f_{1}\left(x_{3}\right) f_{2}\left(x_{1}\right) \quad f_{2}\left(x_{2}\right) \quad f_{2}\left(x_{3}\right) f_{3}\left(x_{1}\right) \quad f_{3}\left(x_{2}\right) \quad f_{3}\left(x_{3}\right)$
$\mid$ 右 $|=|$ 左 {
$f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right) f_{3}\left(x_{3}\right)-f_{1}\left(x_{1}\right) f_{2}\left(x_{3}\right) f_{3}\left(x_{2}\right) f_{1}\left(x_{2}\right) f_{2}\left(x_{3}\right) f_{3}\left(x_{1}\right)-f_{1}\left(x_{2}\right)$

lend{aligned $}$
$\$ \

Heisenberg 和 Dirac 使用（参见第十三章）。交替函数对于构造反对称波函数很重要。

## 物理代写|量子力学代写quantum mechanics代考|Properties of determinants and the Slater determinant

(ii) 矩阵的行列式， $\operatorname{det}(\mathbf{A})$, 给出带有减号的相同行列式， $\operatorname{det}(\hat{A})$, 手术后(Ei) $\leftrightarrow(\mathrm{Ek})$ ，和 $i \neq k$. 这对应于费米子的反对称性质（上一节) 。那是: $\operatorname{det}(\boldsymbol{A})=-\operatorname{det}(\hat{A})$ ，在 哪里 $\hat{\mathrm{A}}:(\mathrm{Ei}) \leftrightarrow(\mathrm{Ek})$
(iii) 如果矩阵的两行或两列具有相同的元素，则其行列式为零。
(iv) 如果运算后得到一个矩阵 $(\lambda \mathrm{Ei}) \rightarrow(\mathrm{Ei})$ ，然后: $\operatorname{det}(\hat{A})=\lambda \operatorname{det}(A)$ ，在哪里 $\hat{\mathrm{A}}:(\lambda \mathrm{Ei}) \rightarrow(\mathrm{Ei})$.
(v) 如果从运算中得到一个矩阵 $(\mathrm{Ei}+\lambda \mathrm{Ek}) \rightarrow(\mathrm{Ei})$ ，和 $\mathrm{i} \neq \mathrm{k}$ ，然后:
$\operatorname{det}(\hat{A})=\operatorname{det}(\boldsymbol{A})$ ，在哪里 $\hat{\mathrm{A}}:(\mathrm{Ei}+\lambda \mathrm{Ek}) \rightarrow(\mathrm{Ei})$. 改变两个轨道后，斯莱特行列式 相同 $(\mathrm{Ei}+\lambda \mathrm{Ek}) \rightarrow\left(\mathrm{E}_{\mathrm{i}}\right)$.
(vi) 如果 $\mathbf{A}$ 和B是相同阶的方阵，那么的衣 $(\boldsymbol{A B})=\operatorname{det}(\boldsymbol{A}) \operatorname{det}(\boldsymbol{B})$
(vii) 矩阵转置的行列式与原矩阵的行列式相同， $\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)$. 然后，MO 轨道可以 在 Slater 行列式的行或列中。

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