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When second quantizating the Schrödinger field

$$\psi(r,t) = \sum_i \phi_i(r)b_i(t),\quad\mbox{and}\quad \psi^{\dagger}(r,t) = \sum_i \phi_{i}(r)^* b_i^{\dagger}(t),$$

we have the commutation relations $[\psi(r,t),\psi^{\dagger}(r',t)]= \delta(r-r')$. Now I want to show that $[b_i,b_j^\dagger] = \delta_{i,j}$.

I tried substituting the expression for $\psi$ into the commutator and got

$$\sum_{i} \sum_{j} \phi_i(r) \phi_j(r')^* [b_i(t), b_j(t)^{\dagger}] = \delta(r-r'),$$ but I don't quite see how this could help me. Does anybody here have an idea how to show this?

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  • $\begingroup$ This is one of those questions where I'd like to answer, but the best solution is for you to struggle with this for a while... $\endgroup$ – Alex Nelson Oct 30 '14 at 19:10
  • $\begingroup$ @AlexNelson this is most often the best way, but don't you have any hint at all? $\endgroup$ – Xin Wang Oct 30 '14 at 19:13
  • $\begingroup$ well, perhaps think about this: what do we expect from $[b_{i}(t), b^{\dagger}_{i}(t)]=??$ What happens if you let $[b_{i}(t), b^{\dagger}_{j}(t)]=f_{ij}(t)$, what properties of $f_{ij}(t)$ should hold (i.e., is it symmetric? Hermitian? Self-adjoint? What's its diagonal?)? What happens when you plug this back in, and use these properties? $\endgroup$ – Alex Nelson Oct 30 '14 at 19:23
  • $\begingroup$ well $[b_i(t),b_i^{\dagger}(t)]$ is self-adjoint. Thus we know that $f_{ii}(t) = f_{ii}^{\dagger}(t)$. Furthermore, we know that $f_{ij}(t) = f_{ji}^{\dagger}(t)$. Not so sure, if that helps me. $\endgroup$ – Xin Wang Oct 30 '14 at 19:44
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The $\phi_i(r)$ form an orthonormal basis of (square-integrable) functions on $\mathbb{R}$, i.e. you should have a relation like $$ \int dr\,\phi_i(r)\phi_j^*(r)=\delta_{ij}. $$ This is what you need in order to expand $\psi(r,t)$ and $\psi^\dagger(r,t)$ the way you did above. You can use this to write the $b_i(t)$ in terms of the $\psi(r,t)$ in the following way: $$ \int dr\,\phi_i(r)^*\psi(r,t)=\int dr\sum_{j}\phi_i^*(r)\phi_j(r)b_j(t)=\sum_j \delta_{ij} b_j (t)= b_i(t). $$ Similarly, you get $$ \int dr\,\phi_i(r)\psi^\dagger(r,t)= b_i^\dagger(t). $$ So, therefore, you have $$ [b_i(t),b_j(t)]=\int dr\,dr'\,\phi_i(r)\phi_j^*(r)[\psi(r,t),\psi^\dagger(r',t)]=\int dr\,dr'\,\phi_i(r)\phi_j^*(r) \delta(r-r')\\ =\int dr\,\phi_i(r)\phi_j^*(r)=\delta_{ij}. $$

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  • $\begingroup$ ah, well nobody told me that they are orthonormal, but thanks for pointing that out. in that case, the exercise is alright :-) $\endgroup$ – Xin Wang Oct 30 '14 at 22:03
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    $\begingroup$ You expand $\psi$ in a basis and upon quantization the expansion coefficients become the ladder operators for the different modes corresponding to the basis function. Often you find the choice of the Fourier basis $\phi_k(t)=e^{ikt}$ which makes the $b_i$ the creation and annihilation operators in the different momentum modes. $\endgroup$ – physicus Oct 30 '14 at 22:06

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