The definition of creation operator for bosonic system is

$$a^{\dagger}|... n_i ...\rangle = \sqrt{n_i + 1} |... n_i + 1 ...\rangle $$

If I take Hermitian adjoint of this I will get $$(a^{\dagger}|... n_i ...\rangle)^{\dagger} = |... n_i ...\rangle^{\dagger} (a^{\dagger})^{\dagger} = \langle... n_i ...| a = \sqrt{n_i + 1} \langle... n_i + 1 ...|$$

But this can't be correct since

$$\langle... n_i ...| a = \sqrt{n_i} \langle... n_i - 1 ...|$$

Can someone explain to me why my Hermitian adjoint equation is wrong.

From the answer I got from JeffDror and the comment left by user26431. I infer that in dual space $A_1 = a$ acts as creation operator and $A_2 = a^{\dagger}$ as annihilation operator.

I am a bit confused by what JeffDror wrote "if you are acting backwards with the operator then you need to use the Hermitian adjoint". In my mind it sounds like that to act with operator A on some bra-vector I must first take Hermitian adjoint of it and than act on bra. It make sense to me if imagine that bra-vector is row vector that I get for transposing ket-vector and non hermitian operator A in ket space is not represented by square matrix. Then in order to multiply matrix by column vector from the left I must transpose matrix that represents A. Is this correct way of thinking about operator how operator A acts in dual space?

  • 1
    $\begingroup$ You have a small typo on your first equation. I think you meant to write, $\sqrt{n _i + 1}\left| ... n _i + 1 ... \right\rangle$ $\endgroup$
    – JeffDror
    Jan 29, 2014 at 15:32
  • $\begingroup$ Annihilation operator acts on bra as creation operator. $\endgroup$
    – user26143
    Jan 29, 2014 at 15:40

1 Answer 1


You're final equation is false. It is not true that, $$ \left\langle ... n _i ... \right| a = \sqrt{ n _i } \left\langle ... n _i - 1 ... \right| $$ The correct result is $$ \left\langle ... n _i ... \right| a = \left\langle ... n _i + 1 ... \right| \sqrt{ n _i +1} $$ as you have just proven.

A non-Hermitian operator doesn't act in the same way to the left as it does to the right. The rules of working with bra's and ket's are if you are acting backwards with the operator then you need to use the Hermitian adjoint (see for example Wikipedia). Otherwise you end up with inconsistencies as you did above. More formally, for any operator $ A$ and states $ \left| \alpha \right\rangle $ and $ \left| \beta \right\rangle $ you can write, \begin{equation} \left\langle \alpha \right| A \left| \beta \right\rangle \equiv \left\langle \alpha |A \beta \right\rangle = \langle A ^\dagger \alpha | \beta \rangle \end{equation}

and only for a Hermitian operator can you omit the dagger.

  • $\begingroup$ I see. But i thought that if $\endgroup$
    – Valjean
    Jan 29, 2014 at 15:38
  • $\begingroup$ ee. But i thought that if $ a |n_i> = \sqrt{n_i} |n_i - 1> $ then $<n_i| a = <n_i-1|\sqrt{n_i}$. That is physics must the same regardless of use ket or bra vektors. For example in Shiff's book on qm he has $-ih \frac{d< \alpha |}{dt} = < \alpha | H^{\dagger} $. Yes hamiltonian is Hermitian here. But does it meter? I pressed enter by mistake in previous comment. $\endgroup$
    – Valjean
    Jan 29, 2014 at 15:51
  • $\begingroup$ I expanded my answer. Hopefully this helps. $\endgroup$
    – JeffDror
    Jan 29, 2014 at 16:08

Your Answer

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

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