TL; DR: According to a book by Tomonaga in a chapter introducing second quantization, the formula $$e^{\pm i \Theta/\hbar} \,\psi(N) = \psi(N\pm 1)$$ is supposed to prove that the second quantization procedure using commutation relations is suitable for bosons. I am able to follow Tomonaga's mathematics up to this formula, but I do not understand how he can conclude that it proves the correctness of his theory for bosons. Could you please explain it to me or point me towards an easy/intuitive explanation? Unfortunately I am not an expert in these topics...

Full question: I am self-studying "The story of spin" by Sin-itiro Tomonaga and I am struggling with lecture 6, where he introduces the second quantization formalism. In this lecture, the author starts from the single particle Schrodinger equation and recast it in terms of canonical equation of motions for Hamiltonian mechanics. Then, he moves to the generalization of these ideas for many particles.

Considering a single particle observable $G$ with eigenvalues $g_n$ each one with a probability $P_n = |a_n|^2$, if $N$ is the number of particles considered then we can write the expectation value $N_n$ for the number of particles of the ensemble with a value $g_n$ for $G$ as $$N_n = NP_n = N |a_n|^2 = A_n A_n^*$$ from which $$ A_n = \sqrt{N} a_n,\quad A_n^* = \sqrt{N} a_n^*. $$ From this he then defines the conjugated momentum as $$ \Pi_n = i \hbar A_n^* $$ from which follows the expectation value of the hamiltonian of the system $$ \bar{H} = -\frac{i}{\hbar} \sum_{n,m} \Pi_n H_{n,m} A_m $$ where $H_{n,m}$ is the matrix element of the single particle hamiltonian. We are now able to derive from these equations the canonical equations for the ensemble $$ \frac{dA_n}{dt} = \frac{\partial\bar{H}}{\partial \Pi_n},\quad\frac{d\Pi_n}{dt} = -\frac{\partial\bar{H}}{\partial A_n} $$ where the following normalization condition holds $$ \sum_n |A_n|^2 = N. $$ At this point, Tomonaga recognizes the important contribution from Dirac to the second quantization formalism, since he redefined $A_n$ and $\Pi_n$ as quantum numbers and he introduced the following commutation relations $$ [A_n,\Pi_n] = i\hbar \delta_{n,m},\quad [A_n, A_m] = [\Pi_n, \Pi_m] = 0. $$ He also states that if we are worried by the fact that $A_n$ and $\Pi_n$ are complex numbers, we can use the following definitions $$ A_n = \sqrt{N_n} e^{i\Theta_n /\hbar},\quad A_n^* = \sqrt{N_n} e^{-i\Theta_n/\hbar} $$ from which we obtain the following hamiltonian $$\bar{H} = \sum_{n,m} \sqrt{N_n} e^{-i\Theta_n /\hbar} H_{n,m} \sqrt{N_m} e^{i\Theta_m/\hbar}. $$ After this mathematics, Tomonaga introduces the concept of "virtual ensembles" and says that in the second quantization formalism we are assuming that we are able to describe such a system of $N$ (non-interacting) particles in terms of the quantum numbers $N_n$ and $\Theta_n$ or $A_n$ and $\Pi_n$. To prove that this is correct, we need to demonstrate that the "usual" description in terms of a wavefunction in the coordinate space satisfying the Schrodinger equation $$ \left[ H(\mathbf{x}_1, \mathbf{p}_1) + H(\mathbf{x}_2, \mathbf{p}_2) + \dots + H(\mathbf{x}_N, \mathbf{p}_N) -i\hbar \frac{\partial}{\partial t} \right] \psi(\mathbf{x}_1, \mathbf{x}_2,\dots,\mathbf{x}_N) = 0 $$ is equivalent to the description in the second quantization formalism $$ \left[\sum_{n,m} \sqrt{N_n} e^{-i\Theta_n/\hbar} H_{n,m} e^{i\Theta_m/\hbar } \sqrt{N_m} -i\hbar \frac{\partial}{\partial t}\right] \psi(N_1, N_2, \dots, N_n,\dots) = 0. $$ To do this, Tomonaga uses an argument by Dirac. He starts considering that $\Theta$ as we have defined it is the momentum conjugated to $N$, so it must hold that $$ \Theta = -i\hbar\frac{\partial}{\partial N} .$$ If this holds, we can make the following series expansion $$ e^{\pm i\Theta/\hbar} = e^{\pm \partial/\partial N} = 1\pm\frac{\partial}{\partial N} \pm \frac{1}{2!}\frac{\partial^2}{\partial N^2} + \dots $$ and applying its right-hand side to the wavefunction and recognizing Taylor's theorem on the series expansion of a function $$ \psi(N) \pm \psi'(N) + \frac{1}{2!} \psi''(N) + \dots = \psi(N\pm 1) $$ we obtain the final equation \begin{equation} e^{\pm i \Theta/\hbar} \,\psi(N) = \psi(N\pm 1).\quad\quad(\star)\end{equation} At this point, Tomonaga's demonstration stops. Quoting his own words:

Since the newly discovered theory has now been proved to be correct for bosons, the use of heuristic theory notwithstanding, there is no reason for us to hesitate to use it. We can proceed with it with confidence. and he starts applying this new formalism.

My problem is that to me is not clear at all why we can conclude from Eq. $(\star)$ that this theory is suitable for bosons and that this approach makes the formulation in the coordinate space using the Schrodinger equation equivalent to the second quantization approach. I am (more or less) able to follow the mathematics, what I miss is probably the physical meaning of Eq. $(\star)$. Can you please help me understanding how is it possible to draw such a conclusion? What is the meaning of Eq. $(\star)$?

It also seems that the very same problem was first discussed in Dirac's paper The Quantum Theory of the Emission and Absorption of Radiation in Eqs. 10 to 13 at pages 251 and 252. In the case of this new demonstration, I am stuck in the demonstration of Eq. 10 starting from the definition in Eq. 8 and the commutation relations reported in the unnumbered equations preceding Eq. 10. Do you have any hint on this?

  • 3
    $\begingroup$ Please do not let posts look like revision histories. The history of a post is publicly accessible for all users, there is no need to reflect it in its newest version and it makes the post harder to read for users reading it for the first time. $\endgroup$
    – ACuriousMind
    Aug 7, 2020 at 8:44

1 Answer 1


Proving that the creation and annihilation operators for bosons exist, namely $$ \begin{equation} e^{\pm i \Theta/\hbar}\psi(N) = \psi(N \pm 1)\end{equation} $$

is one part of a proof by induction. Once we show that the equation produces this result, we need only show that it is true for one special case of N. Tomonaga doesn't explicitly do this... he just assumes you get it.

I believe that Tomonaga is going for a proof by induction. A proof by induction requires proving the equation is valid for a base case, in this case N=1 and then showing that given case N is true, the N+1 case follows. This implies that starting with the wavefunction for N=1 the wavefunction for any N can be obtained.

In this instance the base case isn't made explicit but is clearly true because the equation is just the single particle Schrodinger equation.
$$ \begin{equation} \psi(x_1) = \psi(1)\end{equation} $$

$$ \begin{equation} H\psi(1) - i\hbar\frac{\partial{\psi(1)}}{\partial{t}} = 0\end{equation} $$

Showing that the left hand side can be manipulated to produce the "recipe" for going from the N particle case to the N+1 particle case is the second part of the proof by induction.

Interestingly the word "induction" is never used... I'm not sure why.

  • $\begingroup$ Thank you for your answer, but I am unable to properly understand it. Could you please elaborate it a little more? I agree with you that the operator you wrote is the creation/annihilation operator in the second quantization formalism and that if we assume it to be valid then my starred equation is meaningful. However, how does it help in showing the equivalence between the "usual" description of a wavefunction in coordinate space satisfying the Schroedinger equation and the second quantization formalism? $\endgroup$
    – JackI
    Aug 19, 2020 at 19:59
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    $\begingroup$ I think the point is that the second quantization formalism equation after the Schrodinger equation is true for the N=1 case by inspection. That's the single particle Schrodinger equation. Then, once the creation operator is demonstrated for all cases N by manipulating the left hand side of the equation, the multi particle case is true by induction. You can get to any N starting from N=1. (He could have used some of these words!) $\endgroup$
    – DrFalcon
    Aug 20, 2020 at 14:42
  • $\begingroup$ This seems to be exactly the point I was missing! If you could add this point to your previous answer, I think it would be the perfect answer to my question $\endgroup$
    – JackI
    Aug 21, 2020 at 19:53

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