# Is this deriviation correct?

our professor leaves an exercise for us, to show that $$\frac \partial{\partial <b_k|} <b_k|O|b_k>=2O|b_k>$$ where O is hermitian (you may wonder what $b_k$ stands for... in fact we are talking about Hartree Fock method and $b_1... b_n$ are trial basis states, orthorgonal and normalized, in the slater determinent)

and the derivative with respect to the ket is defined as $$\frac \partial {\partial<b_k|}=[\matrix {\frac \partial {\partial Re(b_k(\vec r,\uparrow)} +i\frac \partial {\partial Im(b_k(\vec r,\uparrow)}\\ \frac \partial {\partial Re(b_k(\vec r,\downarrow)} +i\frac \partial {\partial Im(b_k(\vec r,\downarrow)} }]$$

$b_k(\vec r, \uparrow)$is the wave function of the electron with spin up. The left one is in dirac notation and the right is Schrodinger notation.

I'm considering the following proof, the result is correct but I'm afraid some steps are not well justified, or I did not fully understand the formalism I wrote ...

$O$ is hermitian, so $$O=\sum_i E_i|a_i><a_i|$$ then go to Schrodinger notation $$<b_k|O|b_k>=\sum E_i<b_k|a_i><a_i|b_k>=[\matrix{\sum E_i\int b_k^*(\vec r,\uparrow)a_i(\vec r,\uparrow)dr\int b_k(\vec r,\uparrow)a_i^*(\vec r,\uparrow)dr \\\sum E_i\int b_k^*(\vec r,\downarrow)a_i(\vec r,\downarrow)dr\int b_k(\vec r,\downarrow)a_i^*(\vec r,\downarrow)dr}]$$

then take the derivative of the spin up component $$\frac{\partial ...}{\partial Re(b_k(\vec r, \uparrow))}= \sum E_i\frac{\partial }{\partial Re(b_k(\vec r, \uparrow))} (\int (Re(b_k(\vec r,\uparrow))-i\cdot Im(b_k(\vec r,\uparrow)))a_i(\vec r,\uparrow)dr\int (Re(b_k(\vec r,\uparrow))+i\cdot Im(b_k(\vec r,\uparrow)))a_i^*(\vec r,\uparrow)dr ) \\ =\sum E_i[a_i(\vec r,\uparrow)\int b_k(...)a_i^*(...)dr+a_i^*(...)\int b_k^*(...)a_i(...)dr]\\\frac{\partial ...}{\partial Im(b_k(\vec r, \uparrow))}= \sum E_i\frac{\partial }{\partial Im(b_k(\vec r, \uparrow))} (\int (Re(b_k(\vec r,\uparrow))-i\cdot Im(b_k(\vec r,\uparrow)))a_i(\vec r,\uparrow)dr\int (Re(b_k(\vec r,\uparrow))+i\cdot Im(b_k(\vec r,\uparrow)))a_i^*(\vec r,\uparrow)dr ) \\ =\sum E_i[-i\cdot a_i(\vec r,\uparrow)\int b_k(...)a_i^*(...)dr+ i \cdot a_i^*(...)\int b_k^*(...)a_i(...)dr]$$ Collect things together the second part of both expression disappear and the first terms accumulates. $$\frac {\partial...} {\partial Re(b_k(\vec r,\uparrow)} +i\frac {\partial ...} {\partial Im(b_k(\vec r,\uparrow)} = 2 \sum E_ia_i(\vec r,\uparrow)\int b_k(...)a_i^*(...)dr\\$$ similiar for the spin down and then go to dirac notation( which I understand is : wavefunction→ket, complex conjugate of wave function→bra, integral→bracket) $$\frac \partial{\partial <b_k|} <b_k|O|b_k>=2\sum E_i |a_i><a_i|b_k>=2O|b_k>$$

That seems to be finished, but I am not clear in two points

1. when taking the derivative of function, I just follow this way $$\frac \partial {\partial f(x)} \int f(x)g(x)dx = g(x)$$ Is this correct?
2. Is the translation between dirac notation and schrodinger notation correct?
• In 1., the differential operator should be $\delta/\delta f$, a functional derivative. – J.G. Apr 5 '18 at 19:17
• except for the formalism, how does the integral disappear? I'm not quite familiar with functional derivative...@J.G. – Shuo Apr 5 '18 at 19:23
• The rules are derivable from the generalisation $\delta f(x)/\delta f(y)=\delta(x,\,y)$ (the right-hand side is a Dirac delta) of the familiar result $\partial x_i/\partial x_j=\delta_i^j$ (the right-hand side is a Kronecker delta). – J.G. Apr 5 '18 at 19:50
• Use \langle and \rangle for the angled brackets in latex, they look much nicer than using < and > – enumaris Apr 5 '18 at 20:23
• Could you edit the header to say what this is about? – Pieter Apr 8 '18 at 21:44

In my opinion the first equation of the question, that is $$\dfrac {\partial \langle b_k|O|b_k \rangle}{\partial \langle b_k|} =2O|b_k\rangle \tag{01}$$ doesn't need all this stuff (Hartree-Fock method, Slater determinant, wave function of the electron with spin up etc). To give a simple example, which could be generalized to give the identity (01) : Suppose that the Hilbert space is $\:\mathbb{R}^{n}$, let a variable $\:n-$vector $\:\mathbf{x} \in \mathbb{R}^{n}\:$ and a linear operator represented by the $\: n \times n \:$ real matrix $\:\mathrm{A}$. Let the usual inner product
$$\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathbf{x} \tag{02}$$ a real scalar. Differentiating $$\mathrm{d}\bigl[\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathbf{x}\bigr]=\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathrm{d}\mathbf{x}+\mathrm{d}\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathbf{x}=\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathrm{d}\mathbf{x}+\left(\mathrm{A}^{\boldsymbol{\top}}\mathbf{x}\right) \boldsymbol{\cdot}\mathrm{d}\mathbf{x}=\Bigl[\left(\mathrm{A}+\mathrm{A}^{\boldsymbol{\top}}\right)\mathbf{x}\Bigr]\boldsymbol{\cdot}\mathrm{d}\mathbf{x} \tag{03}$$ where $\:\mathrm{A}^{\boldsymbol{\top}}$ the transpose of $\mathrm{A}$. So formally $$\dfrac{\mathrm{d}\bigl[\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathbf{x}\bigr]}{\mathrm{d}\mathbf{x}}=\left(\mathrm{A}+\mathrm{A}^{\boldsymbol{\top}}\right)\mathbf{x} \tag{04}$$ For real symmetric matrix $\:\mathrm{A}^{\boldsymbol{\top}}=\mathrm{A}$, so $$\dfrac{\mathrm{d}\bigl[\left(\mathrm{A}\mathbf{x}\right) \boldsymbol{\cdot}\mathbf{x}\bigr]}{\mathrm{d}\mathbf{x}}=2\mathrm{A}\mathbf{x} \tag{05}$$
$$\langle\mathrm{A}\mathbf{x},\mathbf{y}\rangle=a_{\imath\jmath}x_{\jmath}\overset{\boldsymbol{\_\!\_}}{y}_{\imath}=x_{\jmath}\overset{\boldsymbol{\_\!\_\!\_\!\_\!\_\!\_\!\_\!\_}}{\overset{\boldsymbol{\_\!\_}}{a}_{\imath\jmath}y_{\imath}}=x_{\jmath}\overset{\boldsymbol{\_\!\_\!\_\!\_\!\_\!\_\!\_\!\_}}{b_{\jmath\imath}y_{\imath}} =\langle\mathbf{x},\mathrm{A}^{\!\boldsymbol{\dagger}}\mathbf{y}\rangle \tag{ed-01}$$ where $\:\mathrm{A}^{\!\boldsymbol{\dagger}}\:$ the hermitian conjugate of $\:\mathrm{A}\:$, that is the complex conjugate of the transposed matrix : $$\mathrm{A}=\{a_{\imath\jmath}\} \quad \Longrightarrow \quad \mathrm{A}^{\!\boldsymbol{\dagger}}=\{b_{\imath\jmath}\}=\{\overset{\boldsymbol{\_\!\_}}{a}_{\jmath\imath}\} \tag{ed-02}$$
• oh.. I see, similar process can be generalized to complex hilbert space and $A^T$ is replaced by $A^\dagger$. But the identity $d(Ax)\cdot x=(A^\dagger x)\cdot dx$ is not obvious to me... so maybe those stuff contains a proof of that. – Shuo Apr 7 '18 at 11:57