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关于密度矩阵的证明题一道

作者 赵端阳
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  • zhihaogong25

    这里面唯一要稍微考虑的的是 rho_{mn}rho_{nm} <= rho_{mm}rho_{nn}, 你可以联系 行列式,
    | rho_{mm}  rho_{mn} |
    | rho_{nm}   rho_{nn} |
    即要这个行列式大于等于0, 然后这是密度矩阵的 一个 余子式, 由于密度矩阵是半正定的, 故等价于 他的一个余子式大于0, 故而这里的
    行列式大于等于0,也就是证明了图上的式子。

  • ra2ghgzh

    第一个不等式,由于[latex]{\rho _{mn}} = \rho _{nm}^*[/latex],所以成立。
    下面证明第二个不等式:
    密度算符为:
    [latex]\rho = \sum\limits_i {{w_i}\left| {{\psi _i}} \right\rangle \left\langle {{\psi _i}} \right|} [/latex]
    因此有:
    [latex]{\rho _{mn}} = \sum\limits_i {{w_i}\left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {{{\psi _i}}} \mathrel{\left | {\vphantom {{{\psi _i}} n}} \right. \kern-\nulldelimiterspace} {n} \right\rangle } [/latex]
    因此有:
    [latex]{\rho _{mm}}{\rho _{nn}} = \sum\limits_{i,j} {{w_i}{w_j}\left( {\left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {{{\psi _i}}} \mathrel{\left | {\vphantom {{{\psi _i}} m}} \right. \kern-\nulldelimiterspace} {m} \right\rangle \left\langle {n} \mathrel{\left | {\vphantom {n {{\psi _j}}}} \right. \kern-\nulldelimiterspace} {{{\psi _j}}} \right\rangle \left\langle {{{\psi _j}}} \mathrel{\left | {\vphantom {{{\psi _j}} n}} \right. \kern-\nulldelimiterspace} {n} \right\rangle } \right)} [/latex]
    互换i和j不会改变上式,因此有:
    [latex]{\rho _{mm}}{\rho _{nn}} = \sum\limits_{j,i} {{w_j}{w_i}\left( {\left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _j}}}} \right. \kern-\nulldelimiterspace} {{{\psi _j}}} \right\rangle \left\langle {{{\psi _j}}} \mathrel{\left | {\vphantom {{{\psi _j}} m}} \right. \kern-\nulldelimiterspace} {m} \right\rangle \left\langle {n} \mathrel{\left | {\vphantom {n {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {{{\psi _i}}} \mathrel{\left | {\vphantom {{{\psi _i}} n}} \right. \kern-\nulldelimiterspace} {n} \right\rangle } \right)} [/latex]
    [latex] = \frac{1}{2}\sum\limits_{i,j} {{w_i}{w_j}} \left( {\left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {{{\psi _i}}} \mathrel{\left | {\vphantom {{{\psi _i}} m}} \right. \kern-\nulldelimiterspace} {m} \right\rangle \left\langle {n} \mathrel{\left | {\vphantom {n {{\psi _j}}}} \right. \kern-\nulldelimiterspace} {{{\psi _j}}} \right\rangle \left\langle {{{\psi _j}}} \mathrel{\left | {\vphantom {{{\psi _j}} n}} \right. \kern-\nulldelimiterspace} {n} \right\rangle + } \right.[/latex]
    [latex]\left. {\left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _j}}}} \right. \kern-\nulldelimiterspace} {{{\psi _j}}} \right\rangle \left\langle {{{\psi _j}}} \mathrel{\left | {\vphantom {{{\psi _j}} m}} \right. \kern-\nulldelimiterspace} {m} \right\rangle \left\langle {n} \mathrel{\left | {\vphantom {n {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {{{\psi _i}}} \mathrel{\left | {\vphantom {{{\psi _i}} n}} \right. \kern-\nulldelimiterspace} {n} \right\rangle } \right)[/latex]

    [latex]{c_{ij}} = \left\langle {m} \mathrel{\left | {\vphantom {m {{\psi _i}}}} \right. \kern-\nulldelimiterspace} {{{\psi _i}}} \right\rangle \left\langle {n} \mathrel{\left | {\vphantom {n {{\psi _j}}}} \right. \kern-\nulldelimiterspace} {{{\psi _j}}} \right\rangle [/latex]
    则有:
    [latex]{\rho _{mm}}{\rho _{nn}} = \frac{1}{2}\sum\limits_{i,j} {{w_i}{w_j}\left( {{{\left| {{c_{ij}}} \right|}^2} + {{\left| {{c_{ji}}} \right|}^2}} \right)} [/latex]
    同理可得:
    [latex]{\rho _{mn}}{\rho _{nm}} = \sum\limits_{i,j} {{w_i}{w_j}{\mathop{\rm Re}\nolimits} \left( {{c_{ij}}c_{ji}^*} \right)} [/latex]
    由于:
    [latex]{\left| {{c_{ij}}} \right|^2} + {\left| {{c_{ji}}} \right|^2} \ge 2{\mathop{\rm Re}\nolimits} \left( {{c_{ij}}c_{ji}^2} \right)[/latex]
    所以第二个不等式正确,

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