1
$\begingroup$

So I am trying to understand in which sense first and second quantization is equivalent. I am following Richard Martin's book on interacting electrons. In the appendix (A.2) he has a piece on second quantization. My understanding of it is that:

First quantization: A state is a square-integrable function.

Second quantization: A state is an infinite-dimensional tuple that simply holds the tally of how many particles are in each energy level of the hamiltonian. Moreover, it seems that second quantization it is assumed that the hamiltonian is for non-interacting harmonic oscillators(?).

In his book on page 680, Richard Martin states that if the matrix elements of an operator $\widehat O(x)$ is going to be the same in first and second quantization, we assume that $\{\psi_n\}$ is a basis of orthonormal functions (are these arbitrary or Harmonic oscilltor?) and that $$ \langle \phi, \widehat O \phi \rangle = \int \phi^* \widehat O \phi\, dx, \quad \text{and} \quad \phi = \sum_n \alpha_n \psi_n, \quad \text{and} \quad |\phi \rangle = \sum_n \alpha_n c_n^\dagger |0 \rangle $$ Martin states that a natural definition is therefore $$ \widehat O = \sum_{k_1,k_2} c_{k_1}^\dagger O_{k_1 k_2} c_{k_2}, \quad O_{k_1 k_2} = \int \psi^*_{k_1} \widehat O \psi_{k_2} dx. $$ What is going on here? Why is this natural?

What is the definition of a summation and scalar multiplication in second quantization? For example, what is the meaning of $$ 5\cdot | 101 \rangle + 42 \cdot |0101 \rangle = ? $$ Without telling the reader what scalar cultiplication and addition means in this hilbert space, how could I tell what $\sum_n \alpha_n c_n^\dagger |0 \rangle$ is? Finally, Martin states that if there is an operator of two arguments $V(x,x')$, then $$ \widehat V = \frac{1}{2} \sum_{k_1k_2k_3k_4} V_{k_1k_2k_3k_4} c_{k_1}^\dagger c_{k_2}^\dagger c_{k_3} c_{k_4}, \qquad V_{k_1k_2k_3k_4} = \int \psi^*_{k_1}\psi^*_{k_2} V \psi_{k_3} \psi_{k_4} dx dx' $$ Why is this the right definition?

$\endgroup$
2
  • $\begingroup$ Instead of an infinity of decoupled oscillators, just consider two. Write and study all expressions for two oscillators, sums supplanting integrals. $\endgroup$ Oct 27, 2020 at 17:38
  • $\begingroup$ Does WP help? $\endgroup$ Oct 27, 2020 at 19:27

1 Answer 1

3
$\begingroup$

Without telling the reader what scalar cultiplication and addition means in this hilbert space

If you're asking about the mathematical meaning, then it is the same as in any other Hilbert space. In fact, mathematicians say that all Hilbert spaces are isomorphic to each other and there's just one Hilbert space.

If you're asking about the physical meaning, then it is still the same as in any other Hilbert space. The interpretation of a state you wrote down is "the system is in a superposition of states $\left| 101 \right>$ and $\left| 0101 \right>$ with real coefficients $5$ and $42$ respectively". You can apply the Born rule to the same extent it can be applied in any other quantum theory and say that the probability of observing $\left| 101 \right>$ is $5^2 / \left( 5^2 + 42^2 \right)$.

Finally, w.r.t. the relationship between a first-quantized operator and the corresponding second-quantized operator. In the Fock space (the second-quantized Hilbert space), a notable subspace is the space of "one-particle states", that is, states with a single occupation number $1$ in some mode and zeroes in all other modes. It is trivial to see that this is the same as the first quantized Hilbert space. Let's denote the mapping from $\mathcal{H}_1$ (first-quantized Hilbert space) to $\mathcal{H}_2$ (second-quantized Hilbert space) by $S$. It is defined by acting on a basis vector in $\mathcal{H}_1$ by $$ S \left| \psi_a \right> = c_a^{\dagger} \left| 0 \right>. $$

The idea is to define the second-quantized opertor corresponding to the first-quantized operator such that it acts in the same way the original operator does, on this subspace. The expression $$ \Omega = \sum_{a,b} c^{\dagger}_a O_{ab} c_b $$ achieves just that (I denote by $O$ the first-quantized operator and by $\Omega$ the second-quantized corresponding operator). It is easy to see: the first-quantized state $\left| \psi_a \right>$ gets mapped into $c_a^{\dagger} \left| 0 \right>$ by the ebmedding of the one-particle subspace in the Fock space. We need to check, the result of applying $O$ gets mapped to the result of applying $\Omega$ to the original operator, that is, we need to check that $S O = \Omega S$. $$ O \left| \psi_a \right> = \sum_b O_{a b} \left| \psi_b \right>, $$ $$ \Omega S \left| \psi_a \right> = \Omega \, c_a^{\dagger} \left| 0 \right> = \sum_{b,c} c_b^{\dagger} O_{bc} c_c c_a^{\dagger} \left| 0 \right> = \sum_{b,c} \delta_{a c} O_{b c} c_b^{\dagger} \left| 0 \right> = \sum_b O_{ab} c_b^{\dagger} \left| 0 \right> = S O \left| \psi_a \right>, $$ which is exactly what we expected to see. Therefore, $\Omega$ acts on the 1-particle subspace of the Fock space exactly like $O$ does on the first-quantized Hilbert space (isomorphic to the 1-particle subspace).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.