# Identities from ket to bra vector

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?

• These questions are all addressed in the definition of a hermitian operator. – user213900 Nov 25 '18 at 19:00
• Are $c_1$ and $c_2$ constants? – exp ikx Nov 25 '18 at 19:03
• They are operators – Elias S. Nov 25 '18 at 19:05
• @Xander while true you comment is not terribly useful. Maybe you could at least augment it with an appropriate link or some additional details. – ZeroTheHero Nov 26 '18 at 0:39
• @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. – user213900 Nov 26 '18 at 8:02

## 2 Answers

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.

• I have edited in question 2. I mean when splitting the inner product into a sum, should I then put a dagger on operators ? – Elias S. Nov 25 '18 at 19:16
• Keep in mind when you take the conjugate transpose of a product, you switch the order of multiplication. – Aaron Stevens Nov 25 '18 at 19:21
• 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. – exp ikx Nov 25 '18 at 19:21
• @Frobenius Yes. Sorry, my bad. I have updated my answer. – exp ikx Nov 26 '18 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: $$\left(\hat c^\dagger\right)^\dagger = \hat c.$$