跳转至

量子多体理论 Quantum many-body theory

1. 又是产生湮灭算符

Fundamental Properties

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

\[ \begin{gathered} \comm{ a }{ a^{\dagger} } = 1 \\ \comm{ a^{\dagger}a }{ a } = -a \\ \comm{ a^{\dagger}a }{ a^{\dagger} } = a^{\dagger} \end{gathered} \]

consider

\[ a^{\dagger}a\ket{\alpha} = \alpha \ket{\alpha} \]

we need \(\alpha \geq 0\)，and we know

\[ ( a^{\dagger}a )( a )^{m}\ket{\alpha} = ( \alpha-m )\ket{\alpha} \]

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

\[ ( a^{\dagger}a )( a )^m \ket{n} = ( n-m )\ket{n} \]

for vacumm state

\[ ( a^{\dagger}a )\ket{0} = 0 \]

we need that the ket is normalized, that we got

\[ \frac{1}{\sqrt{ n! }}( a^{\dagger} )^n \ket{0} =\ket{n} \]

and then we got the eigenvalue of upper&lower operator

\[ a^{\dagger}\ket{n} = \sqrt{ n+1 }\ket{n+1} \qc a\ket{n} = \sqrt{ n }\ket{n-1} \]

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

\[ H = \hbar \omega a^{\dagger}a \]

just for harmonic oscilator, we define

\[ a = \frac{1}{\sqrt{ 2 }}( \hat{X} + i\hat{P} )\qc a^{\dagger} = \frac{1}{\sqrt{ 2 }}( \hat{X} - i\hat{P} ) \]

corresponding harmitonian

\[ H\ket{n} = \hbar \omega( n+\frac{1}{2} ) \]

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

\[ \comm{ q_i }{ p_i } = i\hbar \delta_{i,j}\qc \comm{ q_i }{ q_j } = \comm{ p_i }{ p_j } = 0 \]

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

\[ \mathbb{ O }^T\mathbb{ O }=1\qc \mathbb{ O }^T\mathbb{ V }\mathbb{ O } = diag( \omega_{1}^{2},\omega_{2}^{2},\cdots ) \]

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

\[ H = \sum_i \qty( \frac{p_i^{2}}{2m} + \omega_{i}^{2}Q_{i}^{2} ) \]

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

and

\[ \comm{ \varphi_x }{ \Pi _y } = i\hbar \delta( x-y ) \]

the harmiltonian

\[ H = \frac{1}{2} \int \dd[ 2 ]{ x } \Pi ^{2}( x ) + \frac{1}{2}\int \dd[ 2 ]{ x }\dd[ 2 ]y\ K( x-y )\varphi( x )\varphi( y ) \]

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

\[ \varphi( q ) = \int \dd{x} \varphi( x )e^{ -iqx } \]
\[ K( q ) = \int \dd{x} K( x )e^{ -iqx } \]

so the original eq transform to

\[ H = \frac{1}{2} \int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \Pi ^{\dagger}( q )\Pi ( q ) + \int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \ K( q )\varphi^{\dagger}( q )\varphi( q ) \]

according to the communicate relation:

...

we have

\[ \begin{gathered} a( q ) = \frac{1}{\sqrt{ 2 }}\qty[ \sqrt{ \frac{\omega_q}{\hbar } }\varphi( q ) + \frac{i}{\sqrt{ \hbar \omega_q }}\Pi ( q ) ] \\ a^{\dagger}( q ) = \frac{1}{\sqrt{ 2 }}\qty[ \sqrt{ \frac{\omega_q}{\hbar } }\varphi( q ) - \frac{i}{\sqrt{ \hbar \omega_q }}\Pi ( q ) ] \end{gathered} \]

finally transform

\[ \begin{align} H & = \frac{1}{2} \int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \Pi ^{\dagger}( q )\Pi ( q ) + \frac{1}{2}\int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \ K( q )\varphi^{\dagger}( q )\varphi( q ) \\ & = \frac{1}{2}\int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \hbar \omega_q \qty[ a^{\dagger}( q )a( q ) + a( q )a^{\dagger}( q ) ] \\ & = \int \frac{\dd[ 2 ]{ q }}{( 2\pi )^{2}} \hbar \omega_q ( a^{\dagger}( q )a( q ) + const. ) \end{align} \]

for a specific mode k

\[ a^{\dagger}( k )\ket{0} = \ket{k} \]

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

\[ a^{\dagger}( k_{1} )a^{\dagger}( k_{2} ) = a^{\dagger}( k_{2} )a^{\dagger}( k_{1} ) = \ket{k_{1},k_{2}} \]

we can say that the operator create a particle. furthermore

\[ \comm{ a_i }{ a_j } = 0\qc \comm{ a^{\dagger}_i }{ a^{\dagger}_j }=0\qc \comm{ a^{\dagger}_i }{ a_j }=\delta_{ij} \]

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

\[ \comm{ a( x ) }{ a^{\dagger}( x' ) }_{\mp } = \delta^{3}( p-p' ) \]

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

\[ \hat{A} = \sum_{\alpha,\beta}A_{\alpha,\beta}^{( 1 )}a^{\dagger}( \alpha )a( \beta ) \]

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 $$ \hat{V} := \sum_{i<j}V^{( 2 )}( x_i,x_j ) $$

评论