I have three questions.

Lets say I have a state $\lvert\psi\rangle = \hat{c}_1^\dagger \lvert 0 \rangle $.

  1. Is the corresponding bra then given by $\langle\psi\lvert= \hat{c}_1 \langle 0\lvert $ ? Is it correct to remove the dagger ?

  2. Lets say I want to calculate the inner product given by: $\langle(\hat{c}_1+\hat{c}_2)\langle 0\lvert \hat{S}\lvert (\hat{c}_1^\dagger+\hat{c}_2^\dagger)\lvert 0\rangle$. Is the following re-writing correct?

  3. $\langle\hat{c}_1\langle 0\lvert \hat{S}\lvert (\hat{c}_1^\dagger+\hat{c}_2^\dagger)\lvert 0\rangle$ + $\langle\hat{c}_2\langle 0\lvert \hat{S}\lvert (\hat{c}_1^\dagger+\hat{c}_2^\dagger)\lvert 0\rangle$? Here I do not need to put dagger, only for constants, is that correct?

  • $\begingroup$ These questions are all addressed in the definition of a hermitian operator. $\endgroup$
    – user213900
    Nov 25, 2018 at 19:00
  • $\begingroup$ Are $c_1$ and $c_2$ constants? $\endgroup$
    – exp ikx
    Nov 25, 2018 at 19:03
  • $\begingroup$ They are operators $\endgroup$
    – Elias S.
    Nov 25, 2018 at 19:05
  • $\begingroup$ @Xander while true you comment is not terribly useful. Maybe you could at least augment it with an appropriate link or some additional details. $\endgroup$ Nov 26, 2018 at 0:39
  • $\begingroup$ @zerothehero Point taken, my sincere apologies to the OP. I should have written an answer. A good reference, with worked examples is QM Demystified, by McMahon, which I would highly recommend for study further along the standard course. $\endgroup$
    – user213900
    Nov 26, 2018 at 8:02

2 Answers 2


1) The sign $\dagger$ is placed on matrices to denote its conjugate transpose. Hence, $(\hat{c}^\dagger)^\dagger = \hat{c}$.

It is correct to write $\langle\psi\lvert= (\hat{c_1}^\dagger \lvert 0\rangle)^\dagger = \langle 0\lvert\hat{c_1} $.

If $c$ is a scalar, then $c^\dagger$ = $c^*$.

2) You need to place $\dagger$ for operators acting on a bra vector.

  • $\begingroup$ I have edited in question 2. I mean when splitting the inner product into a sum, should I then put a dagger on operators ? $\endgroup$
    – Elias S.
    Nov 25, 2018 at 19:16
  • $\begingroup$ Keep in mind when you take the conjugate transpose of a product, you switch the order of multiplication. $\endgroup$ Nov 25, 2018 at 19:21
  • $\begingroup$ I am not sure if I understand your 3rd question correctly. The inner product can be split into sums and you don't need to operate dagger just because of splitting. $\endgroup$
    – exp ikx
    Nov 25, 2018 at 19:21
  • $\begingroup$ @Frobenius Yes. Sorry, my bad. I have updated my answer. $\endgroup$
    – exp ikx
    Nov 26, 2018 at 8:46

The notation $\langle\psi\lvert= \hat{c_1} \langle 0\lvert $ is meaningless and should not be used. The correct way to think about your initial state $\lvert\psi\rangle = \hat{c}_1^\dagger \lvert 0 \rangle $ is as a matrix $\hat{c}_1^\dagger$ acting on the column vector $\lvert 0 \rangle$. Switching from a ket to a bra is equivalent to taking the complex-conjugate transpose, and since the transpose reverses the order of multiplication, the correct conjugate is $$\langle\psi\lvert= \langle 0\lvert \hat{c}_1 .$$

Similarly, the notation $$\langle(\hat{c}_1+\hat{c}_2)\langle 0\lvert \hat{S}\lvert (\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle$$ is meaningless and should not be used. The object you're after, if I understand you correctly, is the expectation value of $\hat S$ in the state $|\psi\rangle = (\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle$, and that's given by $$\langle 0\lvert (\hat{c}_1+\hat{c}_2) \hat{S}(\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle.$$ Note that in Dirac notation we never use intermediate bars between operators: it's $\hat{S} (\hat{c}_1^\dagger + \hat{c}_2^\dagger)$, not $\hat{S}\lvert (\hat{c}_1^\dagger + \hat{c}_2^\dagger)$.

Once you've done that, then it's perfectly acceptable to split up any sums, such as e.g. $$\langle 0\lvert (\hat{c}_1+\hat{c}_2) \hat{S}(\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle = \langle 0\lvert \hat{c}_1 \hat{S}(\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle +\langle 0\lvert \hat{c}_2 \hat{S}(\hat{c}_1^\dagger + \hat{c}_2^\dagger)\lvert 0\rangle.$$

As for this,

Here I do not need to put dagger, only for constants, is that correct?

I have no idea what you mean by "only for constants" - there are no constants in your expression. Any operator that's present as its hermitian conjugate should have a dagger; but double-daggers cancel out, in the sense that the hermitian conjugate of a hermitian conjugate returns you to where you started: $$ \left(\hat c^\dagger\right)^\dagger = \hat c. $$


Your Answer

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

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