I was reading "Quantum Optics" by Walls and Milburn, and in chapter 10.1 describing about atom-radiation interaction, it says normal QM equation for single electron in an atom (here $\psi_j(x)$ is an energy eigenstates),

$$\psi(x)=\sum_j a_j \psi_j (x) $$

could be quantized to a field operator, $$\hat{\psi}^\dagger (x)=\sum _j \hat{a_j} ^\dagger \psi^*_j (x) $$ Here $\hat{a_j}^\dagger$ is a creation operator for Fock state.

For normal QM equation, probability for energy state $\psi _j (x)$ being selected is $|a_j|^2$, by Born rule, and probability that this state measured to have specific position $x$ is $|\psi_j (x)|^2$. Therefore probability that $\psi(x)$ is being measured to have energy state $j$ and position $x$ should be $|a_j|^2|\psi_j(x)|^2$.

Now for the case of QFT, field operator $\hat{\psi}^\dagger(x)$ acts on vacuum state $\left|0\right>$, then $$\hat{\psi} ^\dagger (x) \left|0\right>= \sum _j \psi^*_j (x) \hat{a_j} ^\dagger \left|0\right> = \sum _j \psi^*_j (x) \left|1_j \right>$$

Now I believe $\left|1_j \right>$ is a state having energy same as QM state $\psi_j(x)$ and having position exactly $x$. Since we are talking about single electron, I believe $\left|1_j \right>$ is same as $\psi_j(x)$ with fixed position $x$. But instead of probability we calculated above, $\left|1_j \right>$ state has probability of only $|\psi_j(x)|^2$.

What am I missing here?

EDIT: After some search, and thanks to the answer by Charles Francis, I have found the solution to my confusion.

The whole confusion was from the notation. Here, $\psi(x)$ is a wavefunction of single electron state, and I thought $\hat{\psi} ^\dagger (x)$ is an operator generating this wavefunction $\psi(x)$ due to the resemblance of the notation, which is incorrect! Actually textbook has unrigorous approach to quantizing process.

From this lecture note or any other reference about second qunatization one could find basis transformation of creation and annihilation operator, given as $$\hat{a}^\dagger _{\tilde{\alpha}} = \sum_\alpha \left< \alpha | \tilde{\alpha} \right> \hat{a}^\dagger _\alpha $$ Here $\{ \left|\alpha\right> \}$ and $\{ \left|\tilde{\alpha}\right> \}$ is some basis of Hilbert space. Putting basis $\{ \left|x\right> \}$ instead of $\{ \left|\tilde{\alpha}\right> \}$ gives $$\hat{a}^\dagger _{x} = \sum_\alpha \left< \alpha | x \right> \hat{a}^\dagger _\alpha= \sum_\alpha \psi_\alpha^* (x) \hat{a}^\dagger_\alpha $$ One can now notice that field operator $\hat{\psi} ^\dagger (x)$ was actually an operator generating one particle at position $x$. Therefore $\hat{\psi} ^\dagger (x) \left|0\right>=\left|x\right>$, not $\left|\psi\right>$.

And as Charles Francis mentioned, creation operator for $\left|\psi\right>$ analogous to the equation $\psi(x)=\sum_j a_j \psi_j (x) $, needs to be defined independent to the definition of $\hat{\psi} ^\dagger (x)$, and that will be $$\hat{\psi} \equiv \sum_j a_j \hat{a_j}^\dagger$$ Of course, $a_j$ is just an probability coefficient, and $\hat{a_j}^\dagger$ is an creation operator for energy state $j$.

  • $\begingroup$ What does "state has probability of only $\lvert \psi_j(x)\rvert^2$" mean? Probability of what? Also, $\psi_j$ is now an operator, taking its absolute value does not make sense anymore. $\endgroup$
    – ACuriousMind
    Commented Apr 4, 2020 at 11:04
  • $\begingroup$ The book says $\psi_j(x)$ is a complex 'function', while $\psi(x)$ is an operator, since it consists of not only $\psi_j(x)$ but also with $a_j$, which is an 'operator'. To make it unambiguous I will put hat on an operator. $\endgroup$
    – Kenny Kim
    Commented Apr 4, 2020 at 11:23
  • $\begingroup$ This is a handwaving comment by an experimental physicist. I think the fields are represented by the solution of the quantum mechanical equation of the particle under study without a potential, the plane wave solution. The number operator acting on the field gives 1 particle , i.e. its wavefunction which can be integrated , and the sum is like a partial wave analysis of the wavefunction that would have been calculated if the potential were known. $\endgroup$
    – anna v
    Commented Apr 4, 2020 at 12:43

1 Answer 1


I have to say I think the notation is horrible. I would not use it this way but it is clear that $$ \left|1_j \right> = \hat{a_j} ^\dagger \left|0\right> $$ and that this is not related to $\psi$. $\psi_j (x)$ are the coefficients of $\hat{\psi} (x)$ in the energy basis.

One cannot think of $\psi_j (x)$ as referring to definite energy and position, because that is forbidden by the uncertainty principle. This can only be treated as a function of position at particular energy.

Part of the problem is that it is common in applied maths to be ambiguous about whether $\psi (x)$ means the function $\psi$ or whether it means the value of that function at the point $x$. Ket notation can be good for removing that ambiguity.

It may help to write your first equation $$ \psi(x) = \langle x|\psi \rangle = \sum_j \langle x| 1_j \rangle \langle 1_j|\psi\rangle = \sum_j a_j(x)\psi_j. $$ I think this shows that the argument is incorrectly placed. The energy eigenstates have wave functions $a_j(x)$, and the $\psi_j$ are coefficients in the energy basis. It can be written $$ \psi = |\psi \rangle = \sum_j | 1_j \rangle \langle 1_j|\psi\rangle = \sum_j a_j\psi_j. $$

when you quantise it, replacing $|1_j\rangle$ with the creation operator and ignoring $|0\rangle$ this becomes $$ \hat\psi^\dagger = \sum_j {\hat a_j}^\dagger\psi_j $$ and since $\psi$ is arbitrary, we can replace it with $\psi(x)$ so as to construct a field operator $$ \hat\psi^\dagger(x) = \sum_j {\hat a_j}^\dagger\psi_j(x) $$

  • $\begingroup$ So do you mean the whole confusion comes from the horrible notation of the book? If I am understanding your comment right, $\psi_j(x)$ in the QM equation (wave function of energy eigenstate) is different from $\psi_j(x)$ in QFT equation (which is function, but work as coefficient of energy basis). If I denote first one as $\psi_j(x)_{QM}$ and second one $\psi_j(x)_{QFT}$, then their relationship will be $\psi_j(x)_{QFT}=a_j \psi_j(x)_{QM}$. Am I on the right track? $\endgroup$
    – Kenny Kim
    Commented Apr 5, 2020 at 9:05
  • $\begingroup$ Yes, the functions are not the same. I have filled in a bit, but tbh I think the method of second quantisation is always confusing. People are more concerned that they end up with formulae which work, rather than how they actually get the formulae. $\endgroup$ Commented Apr 5, 2020 at 11:16
  • $\begingroup$ Thanks to your answer, and some search by my own, I was able to get solution to this problem. I have edited my question to clarify my confusion once more and added the points you have made in your answer. As from the beginning, it was all notation problem. $\endgroup$
    – Kenny Kim
    Commented Apr 5, 2020 at 15:13
  • $\begingroup$ Well done. Much better to have worked through your own solution. I was wondering about writing down $\hat{\psi} ^\dagger (x) \left|0\right>=\left|x\right>$, but I wasn't exactly sure of your application. $\endgroup$ Commented Apr 5, 2020 at 15:27
  • $\begingroup$ That line came from the chapter 4 of the lecture note I have linked, but if you think it has a problem, I would be pleased to hear your opinion. $\endgroup$
    – Kenny Kim
    Commented Apr 5, 2020 at 15:35

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