Consider the annihilation and raising operators as follows:

$$\hat a|n\rangle=\sqrt{n}|n-1\rangle\qquad\text{and}\qquad\hat a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$

I know normally if I have an operator A then I can write:

$$\langle \phi_j|A| \phi_i\rangle =a_i \langle \phi_j|\phi_i\rangle$$ and for the transpose conjugate we get:

$$\langle \phi_j|A^{\dagger}| \phi_i\rangle =a_j \langle \phi_j|\phi_i\rangle$$ This is to say that the operator acts on the ket whilst the tranpose acts on the bra.

Why is it that with both annihilation and raising operator they both act on the ket and not the bra since the dagger means that they are tranpose conjugates of one another.


2 Answers 2


Abandoning the "sandwich" notation for a moment, one has that $$\langle \phi_i ,A\phi_j\rangle = \langle A^\dagger \phi_i,\phi_j\rangle$$ by the definition of the hermitian conjugate. Both $A$ and $A^\dagger$ are operators on the Hilbert space, but they are related to one another via the inner product. For example, we would also have $$\langle \phi_i,A^\dagger \phi_j\rangle = \langle (A^\dagger)^\dagger \phi_i,\phi_j\rangle = \langle A\phi_i,\phi_j\rangle$$

where we've assumed that $(A^\dagger)^\dagger = A$ (strictly, this is not true for all operators, but it is for the raising/lowering operators on their mutual domain of definition - I am sweeping all such technicalities under the rug).

Having established this, we can re-introduce the sandwich notation by writing $$\langle \phi_i,A\phi_j\rangle \equiv \langle \phi_i | A | \phi_j \rangle \equiv \langle A^\dagger \phi_i , \phi_j\rangle$$

where we can freely interpret the middle expression as the one on the left or the right, i.e. $A$ can either act from the left (in which case it acts as $A$ on $\phi_j$) or from the right (in which case it acts as $A^\dagger$ on $\phi_i$).


Manipulation of combinations of Bras and operators in Dirac notation works like this, $$\begin{aligned} \langle \hat O v | &= |\hat Ov\rangle^\dagger \\ &= (\hat O |v\rangle)^\dagger \\ &= ( |v\rangle )^\dagger \ (\hat O)^\dagger \\ &= \langle v| \hat O^\dagger \end{aligned}$$

Applying this to the ladder operators leads to $$\begin{aligned} \langle n | \hat a^\dagger &= (|n\rangle)^\dagger \hat a^\dagger \\ \langle n | \hat a^\dagger&= \big (\hat a|n\rangle \big)^\dagger\\ \langle n | \hat a^\dagger&= \big (\sqrt n|n-1\rangle \big)^\dagger\\ \langle n | \hat a^\dagger&= (\sqrt n)^* \ \langle n-1| \\ \langle n | \hat a^\dagger&= \sqrt n \ \langle n-1| \ \ \forall n\in \Re \end{aligned}$$

More generally, the equality $$ \langle \hat O^\dagger v | = \langle v| \hat O $$ enables two interpretations when we look at matrix elements like $$\langle v|\hat O |n\rangle. $$ We can associate the operator with the Ket and see it as $$\langle v| \ \big (\hat O |n\rangle\big) $$ , i.e. the operator $\hat O$ acting on the Ket first and then taking the inner product. But equally, we can look at the matrix element as $$\big ( \langle v| \hat O \big) \ |n\rangle =\langle \hat O^\dagger v|n\rangle $$ where the operator is evaluated with the Bra by using the adjungated operator, before we calculate the inner product.

If the operator is self-adjungated, the expressions simplify and you can drop the adjungation of the operator in all steps. Let $\hat H = \hat H^\dagger $, then $$ \langle \hat H v| = \langle v|\hat H ^\dagger = \langle v|\hat H $$

Operators with this property can then be "moved around" in matrix elements as we like, $$ \langle v|\hat H|n\rangle = \langle \hat H^\dagger v|n\rangle = \langle \hat H v|n\rangle $$ This allows you to evaluate self-adjungated operators to the "left or to the right" without the need to know how the adjungated operator acts on your chosen basis.

Response to to question in comment

First of all, the ladder operators are not hermitian/self-adjungated. That means your evaluation was wrong. We have, $$\begin{aligned} \langle n'|\hat a ^\dagger |n\rangle &= \langle n'|\hat a ^\dagger n\rangle\\ &=\sqrt{n+1}\langle n' |n+1\rangle \\[1.0em] \langle n'|\hat a ^\dagger |n\rangle&= \langle \hat a n'|n\rangle \\ &= \sqrt{ n'}^*\langle n'-1|n\rangle \\ &=\sqrt{ n'}\langle n'-1|n\rangle \ \ \forall \sqrt{n'}\in \Re\\[1.0em] \langle n'|\hat a ^\dagger n\rangle &= \langle \hat a n'| n\rangle\\ \sqrt{n+1}\langle n'|n+1\rangle &=\sqrt{ n'}\langle n'-1|n\rangle \\ \end{aligned}$$

You can use both ways and both results are identical. They might look different but if you do the actual numerical evaluation, the value has to be the same. There is not one correct "direction". Both ways are correct and identical. This fact is commonly used to proof that the eigenvectors of hermitian operator are orthogonal.

For example, the number operator $\hat N = \hat a^\dagger \hat a$ is a hermitian operator. We have $$\begin{aligned} \langle n'| \hat N |n\rangle &= n\langle n' |n\rangle\\ &= \langle \hat N^\dagger n' |n\rangle \\ &= \langle \hat N n' |n\rangle \\ &= n'\langle n' |n\rangle \\ \langle n'| \hat N |n\rangle - \langle \hat N n' |n\rangle &= 0\\ (n-n')\langle n'|n\rangle &= 0\\ \end{aligned}$$ Since $$n-n'\neq 0$$ for two different values, we must have $$\langle n'|n\rangle = 0 \ \forall n\neq n' $$.

This proof used the fact that the matrix element can be evaluated in "both directions". You don't have to decide whether you act on the bra or the ket, you can do both as long as you apply the rules properly.

  • $\begingroup$ But suppose we have a situation such as $\langle n'|a^{\dagger}|n\rangle$ how would I know whether this simplifies to $\sqrt{n+1} \langle n'|n+1\rangle$ or to $\sqrt{n'+1} \langle n'+1|n\rangle$ $\endgroup$
    – DJA
    Apr 10, 2021 at 4:11
  • $\begingroup$ @DJA i have updated the answer. $\endgroup$
    – Hans Wurst
    Apr 10, 2021 at 8:52

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