量子多体理论 Quantum many-body theory¶

2026-09-202026-10-16

1. 又是产生湮灭算符¶

Fundamental Properties¶

given \(a\) as a non-hermitian operator.

consider

we need \(\alpha \geq 0\)，and we know

we need \(\alpha-m\geq 0\), so \(\alpha\) must be integer. We redefine it as \(n\) and represents number of particles

for vacumm state

we need that the ket is normalized, that we got

and then we got the eigenvalue of upper&lower operator

for \(a\ket{0} = 0\)，we also have $$ a\ket{0} = \frac{1}{\sqrt{ 2 }} \qty[ x+\pdv{ }{ x } ]\ket{0} =0 $$

finally we define the hamiltonian

just for harmonic oscilator, we define

corresponding harmitonian

for many HOs cases, we define $$ H = \sum_i \frac{p_i^{2}}{2m} + \frac{1}{2}\sum_{i,j} \nu _{ij} \hat{q}_i\hat{q}_j $$

where they movement are independent

if the potential matrix \(\mathbb{ V}\) is positive-definite, we may use an unitary matrix \(\mathbb{ O}\) to diagnize

\(\mathbb{ O}\) transform position \(q\) to norm mode \(Q\), that we go back to uncoupled case:

for continuous space (field quantization), for example a drum head. We transform into continuous arguments $$ \hat{ Q }_i, \hat{ P }_i \implies \varphi_x, \Pi _x $$

and

the harmiltonian

where we assume the elastic function \(K( x)\) satisfied as positive-definite symmetric matrix: $$ K( -x ) = K( x )\qc K( q ) > 0 $$

this time we do Fourier transform

so the original eq transform to

then $$ H = \frac{1}{2} \int \frac{ \dd[2]{q} }{ ( 2\pi )^{2} } \qty[ \Pi ^{\dagger}( q )\Pi( q ) + \omega^{2}( q )\phi ^{\dagger}( q )\phi( q ) ] $$ according to the communicate relation:

.

we have

finally transform

for a specific mode k

for many particle system, we know that normal modes are acting independently. That is the operators are commute

we can say that the operator create a particle. furthermore

we call that particle as Bosan particles.

将平移算符作用在 \(\varphi( x)\) 上

$$ \cdots $$ ...

2¶

for indistinguishable particles, for boson case $$ \Psi_n( x_{1},\cdots ,x_{N} ) = \frac{1}{ \sqrt{ N! }} \sum_P P[ \varphi_{n_{1}}( x_{1} )\cdots \varphi_{n_N}( x_N ) ] $$ for fermion case $$ \Psi_n( x_{1},\cdots ,x_{N} ) =\frac{1}{ \sqrt{ N! }} \sum_P ( - )^{A_P} P[ \varphi_{n_{1}}( x_{1} )\cdots \varphi_{n_N}( x_N ) ] $$ consider norm of the total WF $$ \abs{ \Psi_n( x_{1},\cdots ,x_{N} ) }^{2} = \cdots 回去zi自己补一下 $$ we can see it's normalize

consider inner product of two n-particle state by $$ \begin{align} & \ip{ \phi_{1},\cdots ,\phi_n }{ \psi_{1},\cdots ,\psi_n } \ & = \sum_P \sum_{p'} \cdots \ & = \sum_P \sum_R \cdots \ & = \sum_R ( \pm 1 )^R \ip{ \phi_{R( 1 )} }{\psi_{1} }\ip{ \phi_{R( 2 )} }{\psi_{2} }\cdots \ip{ \phi_{R( n )} }{\psi_{n} } \end{align} $$ we can represent this in determinant $$ \ip{ \phi_{1},\cdots ,\phi_n }{ \psi_{1},\cdots ,\psi_n } =\mdet{ \ip{ \phi_{1} }{ \psi_{1} } & \ip{ \phi_{1} }{ \psi_{2} } & \cdots & \ip{ \phi_1 }{ \psi_n } \ \vdots & \vdots & \ddots & \vdots \ \ip{ \phi_n }{ \psi_{1} } & \ip{ \phi_n }{ \psi_{2} } & \cdots & \ip{ \phi_n }{ \psi_n } }\zeta $$ then consider a boson state, that for any state i can bu \(n_i\) occupied $$ \frac{1}{\sqrt{ n $$ the normalize term is due to diagonal block matrix of its norm. $$ \cdots $$ like project operator, we have $$ 1 = \frac{1}{n!}\sum_{\tilde{\alpha}}!n_{2}!\cdots n_m! }} \ket{\psi_{1},\psi_{1},\cdots ,\psi_{1},\psi_{2},\cdots ,\psi_{2},\cdots ,\psi_{m},\cdots ,\psi_m{1}}\cdots \sumn}\op{ \tilde{\alpha} $$ for fermion case, easily prove from ally it on another state $$ \cdots $$}, \cdots ,\tilde{\alpha}_n

for an general state of multiparticle space $$ \ket{\psi} = \ket{\psi^{( 0 )}} + \ket{\psi^{( 1 )}} + \cdots +\ket{\psi^{( n )}} + \cdots $$ where \(\ket{\psi^{( n )}}\) is an \(n\) particle state.

we can def the creation operator $$ a^{\dagger}( \phi )\ket{\psi_{1},\psi_{2},\cdots ,\psi_n} := \cdots $$ destruction operator $$ a( \phi ) \ket{\psi_{1},\psi_{2},\cdots ,\psi_n} := ( n-1 )-partical state $$ lets find the effect of destruct operator $$ \begin{align} & \mel{ \chi_{1},\cdots ,\chi_{n-1} }{ a( \phi ) }{ \psi_{1},\cdots ,\psi_n } \ & = ( \ip{ \psi_{1},\cdots ,\psi_n }{ \phi,\chi_{1},\cdots ,\chi_{n-1} } )^{*} \ & = judadematrix \ & = \sum_{k = 1}^{n} ( \pm 1 )^{k-1} \ip{ \phi }{ \psi_k }\ip{ \chi_{1},\chi_{2},\cdots ,\chi_{n-1} }{ \psi_{1},\cdots ,\psi_{k-1},\psi_{k+1},\cdots ,\psi_n } \end{align} $$ the chi bra is not related to k, that means $$ a( \phi )\ket{\psi_{1},\cdots ,\psi_{n}} = \sum_{k = 1}^{n} ( \pm 1 )^{k-1} \ip{ \phi }{ \psi_k }\ket{ \psi_{1},\cdots ,\psi_{k-1},\psi_{k+1},\cdots ,\psi_n } $$

easily derive that $$ a^{\dagger}( \phi_{1} )a^{\dagger}( \phi_{2} ) = ( \pm 1 ) a^{\dagger}( \phi_{2} )a^{\dagger}( \phi_{1} ) $$ in this case we have $$ \comm{ a^{\dagger}( \phi_{1} ) }{ a^{\dagger}( \phi_{2} ) }_{\mp } = 0 $$ for bosons are commuted, but fermions are anti-conmuted

now we consider comm relation of create and distruct OP $$ \begin{align} & a( \phi_{1} )a^{\dagger}( \phi_{2} )\ket{\psi_{1},\cdots ,\psi n} \ & = a( \phi \ & = \ip{ \phi_{1} }{ \phi_{2} }\ket{\psi_{1},\cdots ,\psi_n} + \cdots \end{align} $$ one can find $$ \comm{ a( \alpha ) }{ a^{\dagger}( \alpha' ) }} )\ket{\phi_{2},\psi_{1},\cdots ,\psi_n{\mp } = \ip{ \alpha }{ \alpha' }=\delta $$ for boson state we can define $$ \ket{n_{1},n_{2},\cdots } = \frac{ \ket{1,\cdots ,1,2,\cdots ,2,\cdots } }{ \sqrt{ n_{1}!n_{2}!\cdots } } $$ where any state can represent in $$ \ket{n_{1},n_{2},\cdots } = \frac{1}{\sqrt{ n_{1}!n_{2}!\cdots }}( a^{\dagger}( 1 ) )^{n_{1}}( a^{\dagger}( 2 ) )^{n_{2}}\cdots \ket{vac} $$ then we can also prove that $$ a^{\dagger}( \alpha )\ket{n_{1},n_{2},\cdots } = \sqrt{ n_\alpha +1}\ket{n_{1},n_{2},\cdots ,n_\alpha+1,\cdots } $$

we can also change basis $$ \comm{ a( p ) }{ a^{\dagger}( p' ) }_{\mp } = ( 2\pi )^{3}\delta( p-p' ) $$

transform them in FT

now we focus on the hamiltonian. Assume there is an op \(A^{( 1 )}\) act on one-particle state $$ \hat{A}^{( 1 )}( x,p )\ket{\psi} = \ket{\psi'} $$ we want to sum over every particle in an multi-partical state $$ \hat{A} := \sum_j \hat{A}^{( 1 )}( x_j,p_j ) $$ we need to deal with a lot of summation. now we def $$ \hat{A}^{( 1 )}{\alpha\beta} = \mel{ \alpha }{ \hat{A}^{( 1 )}( x,p ) }{ \beta } $$ now we have $$ \hat{A}^{( 1 )}( x,p ) = \sum $$ now the total OP become $$ fkufkufkufkufkufukufku $$} \ket{\alpha}\mel{ \alpha }{ A^{( 1 )} }{ \beta }\bra{\beta} = \sum_{\alpha,\beta}A_{\alpha\beta}^{( 1 )}\op{ \alpha }{ \beta

for a first case we consider an identity operator $$ A^{( 1 )}=\mathbb{ 1 } $$ for definition we have $$ \begin{align} \hat{A} & = \sum_{\alpha,\beta}A_{\alpha,\beta}^{( 1 )}a^{\dagger}( \alpha )a( \beta ) \ & = \sum_{\alpha,\beta}\delta_{\alpha,\beta}a^{\dagger}( \alpha )a( \beta ) \ & = \sum_\alpha a^{\dagger}( \alpha )a( \alpha ) \end{align} $$ next consider the momentum OP $$ P^{( 1 )} = \int \frac{ \dd[ 3 ]{ p } }{ ( 2\pi )^{3} }P\op{ P }{ P } $$ the total momentum

...

Finally consider Hamiltoian $$ \hat{H} = \sum_{i = 1}^{n} -\frac{ \laplacian _i }{ 2m }+V( x_i ) $$ so $$ \begin{align} H & = \mel{ x }{ H }{ x' } \ & = \int\dd[ 3 ]{ x }\dd[ 3 ]{ x' } a^{\dagger}( x )\qty[ \cdots ]a( x' ) \ & = \int\dd[ 3 ]{ x } a^{\dagger}( x )\qty[ \cdots ]a( x ) \end{align} $$

then we consider interaction case. Assume that there are only 2-partical interactions $$ \hat{V} := \sum_{i<j}V^{( 2 )}( x_i,x_j ) = \frac{1}{2}\sum_{i\neq j} V( x_{i},x_{j} ) $$ now we prove that: $$ V = \frac{1}{2} \int\dd[3]{x}\int\dd[3]{y}V^{( 2 )}( x,y )a^{\dagger}( x )a^{\dagger}( y )a( y )a( x ) $$ for the first two operator $$ \begin{align} & a( y )a( x )\ket{x_{1},\cdots ,x_{n}} \ & = a( y )\sum_{k = 1}^{n} ( \pm 1)^{k-1}\delta3 ( x-x_k )\ket{x_{1},\cdots ,x_{k-1},x_{k+1},\cdots ,x_n} \ & = \sum_{k = 1}^{n} ( \pm 1)^{k-1}\delta3 ( x-x_k )\sum_{j = 1\neq k}^{n} \eta_{jk}\delta^3 ( x-x_j ) \ & \ket{x_{1},( \text{ no \(x_j\) \(x_k\) } ),\cdots ,x_{n} } \end{align} $$ where \(\eta = ( \pm 1 )^{j-1}\) if \(j<k\) else \(\eta=( \pm )^j\)

now apply two creation op $$ \begin{align} & a^{\dagger}( x )a^{\dagger}( y )a( y )a( x )\ket{x_{1},\cdots ,x_{n}} \ & = \cdots \ & = \sum_{j\neq k} \delta^3( x-x_k )\delta^3( y-x_j )\ket{x_{1},\cdots ,x_{n}} \end{align} $$ integrate over whole space and we find it's true

now we adding new spin term $$ a^{{\dagger}{\sigma}( x )\ket{vac} = \ket{( x,\sigma )} $$ then we can counter interaction with different spin $$ V = \frac{1}{2} \int\dd[3]{x}\int\dd[3]{y}V^{( 2 )}( x,y )a}( y )a_\sigma( x ) $$ do FT $$ a_\sigma( x ) = \int\frac{ \dd[3]{k} }{ ( 2\pi )^{3} }e^{ -ikx }a_{\sigma}( k ) $$ and $$ a^{\dagger}}( x )a_{\sigma'}^{\dagger}( y )a_{\sigma'\sigma( x ) = \int\frac{ \dd[3]{k} }{ ( 2\pi )^{3} }e^{ ikx }a^{\dagger}( k ) $$ then the potential term becomes $$ \begin{align} V( q ) & = \int\dd[3]{( x-y )}V( x-y )e^{ -iq( x-y ) } \ & = \frac{1}{2}\sum_{\sigma,\sigma'} \int \frac{ \dd[3]{q} }{ ( 2\pi )^{3} } \int \frac{ \dd[3]{k} }{ ( 2\pi )^{3} } \int \frac{ \dd[3]{k'} }{ ( 2\pi )^{3} } \ & V( q ) a_{\sigma'}^{\dagger}( k+q )a_{\sigma'}^{\dagger}( k'-q )a_{\sigma'}( k' )a_\sigma( k) \end{align} $$ we can draw it by Fermon Graph

...

Jellium model¶

Jellium model is an effective model for a simple metal. We can think it as electrons move in a positive charge background. Several assumptions:

1. \(+Z\) cation to be static

2. uniformly distributed positive-charge background $$ n( x ) = n = \frac{N}{V} $$

3. the size of the box is large enough.

and we consider no Bloch Bond theory.

consider the e-e interaction hamiltonian $$ \begin{align} H_{el} & = \frac{1}{2}\int\dd[3]{x}\int\dd[3]{y}a_{\sigma}^{\dagger}( x )a_{\sigma'}^{\dagger}( y )\frac{ e^{2} }{ | x-y | }a_{\sigma'}( y )a_{\sigma}( x ) \ & = \frac{1}{2}\int\dd[3]{x}\int\dd[3]{( x-y )} V( x-y )\qty[ N^{2}-N ] \ & = \frac{1}{2} \qty( \frac{1}{V} )^{3} \sum_{q\neq 0}\sum_{k}\sum_{k'} \frac{ 4\pi e^{2} }{ \mu^{2}+q( k' )a_\sigma( k) \end{align} $$ b-b interaction $$ H_{b} = \frac{e^{2}}{2} \int\dd[3]{x}\int\dd[3]{y} \frac{ n( x )n( y ) }{ | x-y | } $$ eee, this one is not converge. Do a little trick: adding a exponent term $$ H_{b} = \frac{e^{2}}{2} \int\dd[3]{x}\int\dd[3]{y} \frac{ n( x )n( y ) }{ | x-y | } e^{ -\mu \abs{ x-y } } $$ then the integration is converge. Finally let } } a_{\sigma'}^{\dagger}( k+q )a_{\sigma'}^{\dagger}( k'-q )a_{\sigma'\(\mu \to 0\) we get back to original formula. $$ \begin{align} H_b & = \frac{e^{2}}{2} \qty( \frac{N}{V} )^{2} \int \dd[3]{x}\int\dd[3]{z} \frac{ e^{ -\mu| z | } }{ | z | } \ & = \frac{e^{2}}{2} \frac{N^{2}}{V} \frac{4\pi}{\mu^{2}} \end{align} $$ we can see the \(H_b\) diverge, which represent long-range nature of Coulomb interaction

e-b interaction $$ \begin{align} H_{el-b} & = -e^{2} \sum_{i = 1}^{N} \frac{N}{V} \int\dd[3]{x} \frac{ e^{ -\mu| x-x_{i} | } }{ | x-x_{i} | } \ & = -e^{2} \frac{N^{2}}{V} \frac{4\pi}{\mu^{2}} \end{align} $$ now just rewrite the e-e interaction

Field theory at 0K¶

(Adiabatic) Gell-Mann Low Theorem¶

consider Hamiltonian $$ H = H_{0} + e^{ -\varepsilon| t | }H_{1} $$ which means the Hamiltonian changes form \(H_{0}\) to \(H_{0}+H_{1}\) when \(t=-\infty \to t=0\).

[Picture: some energy level transfer to another one, without energy level crossing]

Using time evolving operator $$ \hat{U}( t,t_{0} )\ket{\Phi_0} = \ket{\Phi( t )} $$ Schrodinger pict $$ ih\pdv{ t } \ket{\Psi_S( t )} = H\ket{\Psi_S( t )} $$ Hinsberg pict $$ i\hbar \pdv{ t } O_H( t ) = \comm{ O_H( t ) }{ H } $$ interaction pict $$ ih\pdv{ t } \ket{\Psi_I( t )} = H_{1}\ket{\Psi_I( t )} $$

so for seconf quant. we def schrodinger operator as $$ O_S = f( a^{\dagger}k, a_k ) $$ the corresponding interaction OP is $$ O_I( t ) = f( e^{ \frac{iH}t}{\hbar } }a^{\daggerke^{ -\frac{iH}t}{\hbar } }, e^{ \frac{iH_{0}t}{\hbar } }a^{\daggerke^{ -\frac{iH ) $$ from interaction pict $$ \ket{\Psi_i( t )} = U( t;t_{0} ) \ket{\Psi_i( t_{0} )} $$ the LH is $$ \ket{\Psi_i( t )} = e^{ iH_{0}t/\hbar } e^{ iH( t-t_{0} )/\hbar }e^{ -iH_{0}t/\hbar } \ket{\Psi_I( t_{0} )} $$ this gives $$ U( t;t_{0} ) = e^{ iH_{0}t/\hbar } e^{ iH( t-t_{0} )/\hbar }e^{ -iH_{0}t/\hbar } $$ by deriverate it by t we have $$ i\hbar \pdv{ t } U( t;t_{0} ) = H_{1}( t )U( t;t_{0} ) $$ integrate it $$ \begin{align} U( t;t_{0} ) - 1 & = -\frac{i}{\hbar } \int_{t_{0}}^{t} \dd{t'} H( t' )U( t';t_{0} ) \end{align} $$ this gives $$ U( t;t_{0} ) = 1-\frac{i}{\hbar }\int_{t_{0}}^{t} \dd{t'} H_{1}( t' ) U( t';t_{0} ) $$ by repeating this we have $$ \begin{align} U( t;t_{0} ) & = 1+( \frac{ -i }{ \hbar } )\int_{t_{0}}^{t} \dd{t'} H_{1}( t' ) \ & + ( \frac{ -i }{ \hbar } )^{2} \int_{t_{0}}^{t} \dd{t'} \int_{t_{0}}^{t'} \dd{t''} H_{1}( t' )H_{1}( t'' ) \ & + \cdots \end{align} $$ the third term we rerwrite $$ \frac{1}{2} \int_{t_{0}}^{t} \dd{t'} \int_{t_{0}}^{t} \dd{t''} \qty[ H_{1}( t' )H_{1}( t'' )\theta( t'-t'' ) + H_{1}( t'' )H_{1}( t' )\theta( t''-t' ) ] $$ this two are not commute. we call it as }t}{\hbar } time ordered product of operator denoteed by Tau $$ \frac{1}{2} \int_{t_{0}}^{t} \dd{t'} \int_{t_{0}}^{t} \dd{t''} \Tau( H_{1}( t' )H_{1}( t'' ) ) $$ then the n-th term can write as $$ \frac{1}{n!} \int_{t_{0}}^{t} \dd{t_{1}}\int_{t_{0}}^{t} \dd{t_{2}}\cdots \int_{t_{0}}^{t} \dd{t_{n}} \Tau\qty[ H_{1}( t_{1} )\cdots H_n( t_n ) ] $$ so we can write that $$ \begin{align} U( t;t_{0} ) & = \sum_{n = 0}^{\infty} ( \frac{ -i }{ \hbar } )^n \frac{1}{n!} \int_{t_{0}}^{t} \dd{t_{1}}\int_{t_{0}}^{t} \dd{t_{2}}\cdots \int_{t_{0}}^{t} \dd{t_{n}} \Tau\qty[ H_{1}( t_{1} )\cdots H_n( t_n ) ] \ & = \Tau\qty{ \exp[ ( \frac{ -i }{ \hbar }\int_{t_{0}}^{t} \dd{t'} H( t' ) ) ] } \end{align} $$

Adiabatic switch¶

consider a new Hamiltonian $$ H = H_{0} + e^{ -\varepsilon| t | }H_{1} $$ evolving from \(t = -\infty\) to \(t=0\)

we obtain $$ \ket{\Psi_I( 0 )} = U_{\varepsilon}( 0-\infty )\ket{\Psi_0} $$ Gell-Manm & Low Theorem Suggest that this state is an exact eigenstate of the interacting system $$ \ket{\Psi_{0}( \varepsilon )} \equiv \frac{ U_{\varepsilon}( 0;-\infty )\ket{\Phi_{0}} }{ \ev{ U_{\varepsilon}( 0;-\infty ) }{ \Phi_{0} } } $$ GL Theorem: Suppose \(\ket{\Psi_{0}( \varepsilon )}\) is expanded in a power of g $$ \ket{\Psi_{0}( \varepsilon )} = \ket{a_{0}( \varepsilon )} + g\ket{a_{1}( \varepsilon)} + g^{2} \ket{a_{2}( \varepsilon )} + \cdots
$$ if the coeffivient \(\ket{a_i( \varepsilon )}\) are all converged in the limit of \(\varepsilon \to 0\) $$ ( H_{0}+H' )\ket{\Psi_0( \varepsilon )} = ( E_{0}+\Delta E )\ket{\Psi_0( \varepsilon )} $$ to prove that first we have $$ ( H_{0} - E_{0} )U_{\varepsilon}( 0,-\infty )\ket{\Phi_0} = \comm{ H_{0} }{ U_{\varepsilon}( 0,-\infty ) }\ket{\Phi_{0}} $$ as we known before $$ U_{\varepsilon}( 0,-\infty ) = 1+ \sum_{n = 1}^{\infty} ( \frac{ -i }{ \hbar } )^n \frac{1}{n!} \int_{t_{0}}^{t} \dd{t_{1}}\int_{t_{0}}^{t} \dd{t_{2}}\cdots \int_{t_{0}}^{t} \dd{t_{n}} \Tau\qty[ H_{1}( t_{1} )\cdots H_n( t_n ) ] $$ let's consider the n-thassociatd commutator $$ \begin{align} \comm{ H_{0} }{ H_I'( t_{1} )H_I'( t_{2} )\cdots H_I'( t_n ) } & = -i\hbar ( \pdv{ t_{1} } + \pdv{ t_{2} } + \cdots + \pdv{ t_n } ) H_I'( t_{1} )H_I'( t_{2} )\cdots H_I'( t_n ) \end{align} $$ this is because $$ -i\hbar \pdv{ t } H_{I}'( t ) = \comm{ H_{0} }{ H_I'( t ) } $$ plug into $$ \begin{align} \comm{ H_{0} }{ U_{\varepsilon}( 0,-\infty ) } & = i\hbar \sum_{n = 1}^{\infty} \frac{1}{n!} ( -\frac{i}{\hbar } )^n \int_{-\infty}^{0} \dd{t_{1}} \cdots \int_{-\infty}^{0} \dd{t_{n}} e^{ \varepsilon( t_{1}+\cdots +t_n ) } \ & \times n\pdv{ t_{1} } \qty{ \Tau\qty[ H_I'( t_{1} )\cdot \cdots \cdot H_{I}'( t_n ) ] } \end{align} $$ this become $$ \begin{align} & -H_I'\ket{\Psi_{0}( \varepsilon )} + \varepsilon\sum_{n = 1}^{\infty} \frac{1}{( n-1 )!} ( -\frac{i}{\hbar } )^{ n-1 } \int_{-\infty}^{0} \dd{t_{1}} \cdots \int_{-\infty}^{0} \dd{t_{n}} e^{ \varepsilon( t_{1}+\cdots +t_n ) }\ & \times n \qty{ \Tau\qty[ H_I'( t_{1} )\cdot \cdots \cdot H_{I}'( t_n ) ] } \ket{\Psi_{0}( \varepsilon )} \end{align} $$ we assume that the H1 is proportional to a coupling constant g $$ H' \to gH' $$ so $$ ( \frac{ -i }{ \hbar } )^{ n-1}\frac{1}{( n-1 )!}g^n = i\hbar g\cdot \pdv{ g } \qty{ ( \frac{ -i }{ \hbar } )^{ n} \frac{1}{n!}g^n } $$ the second term becom $$ i\hbar \varepsilon g\pdv{ g } \ket{\Psi_0( \varepsilon )}
$$ finnally that $$ ( H_{0}-E_{0} )\ket{\Psi_{0}( \varepsilon )} = -H'\ket{\Psi_0( \varepsilon )} + i\hbar \varepsilon g\cdot \pdv{ g } \ket{\Psi_o( \varepsilon )} $$ multipli it by \(\displaystyle \frac{ \bra{\Phi_{0}} }{ \ip{ \Phi_{0} }{ \Psi_{0}( \varepsilon ) } }\)

we have $$ \frac{ \mel{ \Phi_{0} }{ H_{1} }{ \Psi_{0}( \varepsilon ) } }{ \ip{ \Phi_{0} }{ \Psi_{0}( \varepsilon ) } } = i\hbar \varepsilon g\cdot \pdv{ g } \ln( \ip{ \Phi_{0} }{ \Psi_{0}( \varepsilon ) } ) $$ 看天书

Green function¶

green funtion is a "probe" of \(H = H_{0}+H'\) at \(T=0\) or low-energy excited state of H $$ iG( x,t;x',t' ) = \frac{ \ev{ \Tau\qty[ \Psi_H( x,t )\Psi_H^{\dagger}( x',t' ) ] }{ \Psi_{0} } }{ \ip{ \Psi_{0} }{ \Psi_{0} } } $$ where \(\ket{\Psi_{0}}\) is the ground state and \(\Psi_H^{\dagger}( x,t )\) is a creation operator in hinsberg pict $$ \Psi_H^{\dagger}( x,t ) = e^{ iHt/\hbar } \Psi ^{\dagger}( x ) e^{ -iHt/\hbar } $$ time order product $$ T( \phi( 1 )\phi(2 )\cdots \phi( n ) ) = \sum_P ( \pm )^P \theta( t_{P_{1}} - t_{P_{2}} )\cdots \theta( t_{P( n-1 )} - t_{P_n}) \phi( P( 1 ) )\cdots \phi( P( n ) ) $$ the RHS becomes $$ \begin{align} & \theta( t-t' ) e^{ iE_{0}( t-t' )/\hbar } \frac{ \mel{ \Phi_{0} }{ \Psi( x )e^{ -iH( t-t' ) }\Psi ^{\dagger}( x' ) }{ \Psi_{0} } }{ \ip{ \Psi_{0} }{ \Psi_{0} } } \ \pm & \theta( t'-t ) e^{ -iE_{0}( t-t' )/\hbar } \frac{ \mel{ \Phi_{0} }{ \Psi ^{\dagger}( x' )e^{ -iH( t-t' ) }\Psi( x ) }{ \Psi_{0} } }{ \ip{ \Psi_{0} }{ \Psi_{0} } } \end{align} $$ introduce 2 single particle operator

..

first we consider non-interacting fermi system $$ H = H_{0} = \int\dd[3]{x} \psi ^{\dagger}( x ) ( -\frac{ \hbar^{2}\grad ^{2} }{ 2m } )\psi( x ) + .. $$ we have fermi-sea state

... $$ \begin{align} \psi_n( x ) & = \ & = \theta( t-t' )\frac{1}{V} \sum_k e^{ ik( x-x' ) - i\varepsilon_k( t-t' ) }\ev{ cc^{\dagger} }{ \psi_0 } \ & - \theta( t'-t )\frac{1}{V} \sum_k e^{ ik( x-x' ) - i\varepsilon_k( t-t' ) }\ev{ c^{\dagger}c }{ \psi_0 } \ & = \theta( t-t' )\frac{1}{V} \sum_k e^{ ik( x-x' ) - i\varepsilon_k( t-t' ) }\theta( | k | - k_F ) \ & - \theta( t'-t )\frac{1}{V} \sum_k e^{ ik( x-x' ) - i\varepsilon_k( t-t' ) }\theta( k_F - | k | ) \ & = \frac{i}{V} e^{ ik( x-x' ) } \int_{-\infty}^{\infty} \frac{\dd{\omega}}{2\pi} e^{ -i\omega( t-t' ) } \qty{ \frac{ \theta( | k | - k_F ) }{ \omega-\varepsilon_k+i \eta } + \frac{ \theta( k_F - | k |) }{ \omega-\varepsilon_k-i \eta } } \end{align} $$ using res $$ G_{0}( k,\omega ) =\frac{ \theta( | k | - k_F ) }{ \omega-\varepsilon_k+i \eta } + \frac{ \theta( k_F - | k |) }{ \omega-\varepsilon_k-i \eta } $$ two singlarity on $$ \omega = \varepsilon_k \pm i \eta $$ 分别对应 \(-E_n( n-1 )+E_{0}\) 和 \(E_n( N+1 ) - E_{0}( N )\)

现在我们研究带相互作用的体系

define sea state momentum $$ \hat{P}\mu = \sum\hbar k\mu c^{\dagger}kc_k $$ then we can obtain a tranposition $$ \psi( x+y ) = e^{ -i\hat{P}y/\hbar }\psi( x )e^{ i\hat{P}y/\hbar } $$ so $$ \psi_H( x ) = e^{ i\hat{H}t/\hbar - iPx/\hbar } \psi( 0 )e^{ -i\hat{H}t/\hbar + iPx/\hbar } $$ then we can separate H $$ \begin{align} \hat{H} & = \int\dd[3]{x} \psi ^{\dagger}( x ) ( -\frac{ \hbar^{2}\grad ^{2} }{ 2m } )\psi( x ) \ & + \int\dd[3]{x}\dd[3]{y} \psi ^{\dagger}( x )\psi ^{\dagger}( y )V( x-y )\psi( x )\psi( y ) \end{align} $$ then green func becomes $$ \begin{align} iG( x,t;x',t' ) & = \frac{ \ev{ \Tau\qty[ \Psi_H( x,t )\Psi_H^{\dagger}( x',t' ) ] }{ \Psi \ & = \frac{1}{\ip{ \Psi_{0} }{ \Psi_{0} }} \qty{ \theta( t-t' )\mel{ 0,N }{ e^{ i\hat{H}t/\hbar -iPx/\hbar }\psi( 0 )e^{ -i\hat{H}( t-t' )/\hbar +i\hat{P}( x-x' )/\hbar }\psi ^{\dagger}( 0 )e^{- i\hat{H}t/\hbar +iPx/\hbar } }{ 0,N } - \cdots } \end{align} $$ then we plug in this projection $$ \sum_n \op{ n,N-1 } = 1 $$ on the left of second } } }{ \ip{ \Psi_{0} }{ \Psi_{0} } \(\psi ^{\dagger}( 0 )\), the term become $$ \theta( t-t' ) \sum_{n \in N+1} e^{ -i( E_n( N+1 ) - E_{0}( N ) )( t-t' )/\hbar }e^{ i( P_n( N+1 ) - P_0( N ) )( x-x' )/\hbar }\cdot | \mel{ 0,N }{ \psi( 0 ) }{ n,N+1 } |^{2} $$ another term same

finally we have $$ \begin{align} G( k,\omega ) & = V\sum_{n \in N+1} \delta_{k,P_n( N+1 )/\hbar } \frac{ | \mel{ 0,N }{ \psi( 0 ) }{ n,k,N+1 } |^{2} }{ \omega - ( E_n( N+1 ) - E_0( N ) )/\hbar +i \eta } \ & + V\sum_{n \in N-1} \delta_{k,P_n( N-1 )/\hbar } \frac{ | \mel{ 0,N }{ \psi ^{\dagger}( 0 ) }{ n,-k,N-1 } |^{2} }{ \omega + ( E_n( N-1 ) - E_0( N ) )/\hbar - i \eta } \end{align} $$ it have inf singularity symmetry distrubute by the side of \mu $$ \omega =E_n( N+1 ) - E_0( N ) = \Delta E_n( N+1 ) + \mu $$ 另外， \(\ket{n,k,N+1}\) 是一个 quasi-particle，也可以说是一个dressed-particle。包括粒子出现在费米海上的态和他所有激发态的叠加 $$ \begin{align} \ket{n,k,N+1} & = c_{k_{1}}^{\dagger}\ket{F.S.} \ & + c_{k_{1}}^{{\dagger}c_{k_{2}}} \ & + \cdots \end{align} $$}c_{k_{3}}\ket{F.S.

Feynman-Dyson perturbation Theory $$ \psi_H( x) = e^{ iHt/\hbar }e^{ -iH_{0}t/\hbar }e^{ iH_{0}t/\hbar }\psi( x )e^{ iH_{0}t/\hbar }e^{ -iH_{0}t/\hbar }e^{ -iHt/\hbar } $$ 其实可以继续化简成这样 $$ \psi_H( x ) = U_{\varepsilon}( 0,t )\psi_I( x )U_{\varepsilon}( t,0 ) $$ g.s. of H0 $$ \ket{\Psi_{0}} = \frac{ U_{\varepsilon}( 0,-\infty )\ket{\Psi_{0}} }{ \ev{ U_{\varepsilon}( 0,-\infty ) }{ \Psi_0 } } $$ inner product can be write as $$ \theta( t-t') \ev{ U( -\infty,t )\psi_I( x )U_{\varepsilon}( t,t' )\psi_I^{\dagger}( x' )U( t',-\infty ) }{ \psi_{0} } \pm h.c. $$ ? $$ U_{\varepsilon}( +\infty,-\infty ) = \Tau\qty{ U_{\varepsilon}( +\infty,t )U_{\varepsilon}( t,t' )U_{\varepsilon}( t',-\infty ) }\theta( t-t' ) $$ ???

经过了一堆天书 $$ \begin{align} U_{\varepsilon}( +\infty,-\infty ) & = \Tau\qty{ \exp[ - \frac{i}{\hbar } \int_{-\infty}^{\infty} \dd{s} H_I'( s ) ] } \ & = \sum_{n = 0}^{\infty} \frac{1}{n!}\qty( -\frac{i}{\hbar } )^{n} \int\dd{s_{1}} \int\dd{s_{2}} \cdots \int\dd{s_n}\Tau\qty( H_I'( s_{1} )\cdots H_I'( s_n ) ) \end{align} $$ we seperate $$ \Psi_I( x ) = \Psi^{( + )}I( x ) + \Psi_I^{( - )}( x ) $$ we have $$ \Psi^{( + )}( x ) $$ left part is a destruction part, right is creation part}( x )\ket{\Phi_{0}} = 0 = \bra{\Phi_{0}} \Psi_I^{( - )

we have $$ \ket{\Phi_{0}} = \prod_{| k |<k_F}c_{k}^{\dagger}\ket{vac} $$

boson case

destruction $$ \psi_I( x ) = \psi_I^{( - )}( x ) $$ creation $$ \psi_I^{\dagger}( x ) = ( \psi ^{\dagger}_I( x ) )^{( + )} $$

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