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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?

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  • $\begingroup$ 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. $\endgroup$
    – ZeroTheHero
    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

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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.

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  • $\begingroup$ 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. $\endgroup$
    – BioPhysicist
    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
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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: $$ \left(\hat c^\dagger\right)^\dagger = \hat c. $$

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