In above formula, I have confused what does mean $\left\langle{\hat{O}}^\dagger\varphi\right|$.

  1. ${\hat{O}}^\dagger$ acts to $\left|\left.\varphi\right\rangle\right.$ first and then operate Hermitian conjugate.
  2. ${\hat{O}}^\dagger$ acts to $\left\langle\left.\varphi\right|\right.$.

Matrix calculation and function calculation had worked different. (I don't know if it's right, but based on what I calculated)

For example, suppose that $\left|\left.\psi\right\rangle\right.=\left[\begin{matrix}1\\0\\\end{matrix}\right],\ \ \left|\left.\varphi\right\rangle\right.=\left[\begin{matrix}1\\1\\\end{matrix}\right],\ \ \hat{O}=\left[\begin{matrix}1&i\\i&2\\\end{matrix}\right]$




So, I thought $\left\langle\left.{\hat{O}}^\dagger\varphi\right|\right.$ means that ${\hat{O}}^\dagger$ acts to $\left|\left.\varphi\right\rangle\right.$ first and then operate Hermitian conjugate.

Suppose that function $f, g$ and ladder operator ${\hat{a}}_\pm$ which are Hermitian conjugate each other.

$\int_{-\infty}^{\infty}{f^*\left({\hat{a}}_\pm g\right)dx}=\frac{1}{\sqrt{2hm\omega}}\int_{-\infty}^{\infty}{f^*\left(\mp\hbar\frac{d}{dx}+m\omega x\right)gdx}$

$\int_{-\infty}^{\infty}{f^*\left(\mp\hbar\frac{d}{dx}+m\omega x\right)gdx}=\int_{-\infty}^{\infty}{\mp\hbar f^*\left(\frac{dg}{dx}\right)dx}+\int_{-\infty}^{\infty}{m\omega xf^*gdx}=\left.\mp\hbar f^*g\right|_{-\infty}^\infty\pm\int_{-\infty}^{\infty}\hbar\left(\frac{df^*}{dx}\right)gdx+\int_{-\infty}^{\infty}{m\omega xf^*gdx}=\int_{-\infty}^{\infty}{\left(\pm\hbar\frac{d}{dx}+m\omega x\right)f^*gdx}$


$\int_{-\infty}^{\infty}{f^*\left({\hat{a}}_\pm g\right)dx}=\int_{-\infty}^{\infty}{{\hat{a}}_\mp f^*gdx}$

Now, ${\hat{a}}_\mp$ acts to $f^*$. According to the matrix calculation I expected $\left({\hat{a}}_\mp f\right)^\dagger$ but it’s different.

  • 1
    $\begingroup$ Since your ladder operators are real, I think you can pull them inside the conjugate, can't you? Then you got your answer. $\endgroup$ Mar 9 at 14:45
  • $\begingroup$ Ladder operator is real, however it's not hermitian operator since (d/dx)^T = -(d/dx) $\endgroup$
    – XX X
    Mar 9 at 15:34

3 Answers 3


If you wish to use Dirac notation then you should not put operators inside either bra or ket symbols. You can write $$ \hat{Q} | \phi \rangle $$ and $$ \langle \psi | \hat{Q} | \phi \rangle $$ and $$ ( \hat{Q} | \phi \rangle )^\dagger = \langle \phi | \hat{Q}^\dagger $$ but the notation $$ \langle \hat{Q}^\dagger \phi | \;\;\;\;\;\;\;\;\;???? $$ is an abuse of Dirac notation. Of course you can write it if you want to adopt some other notation: that is up to you. You could, for example, announce that this quantity shall be defined, in your own work, to be identical to $$ \langle \phi | \hat{Q} $$ which appears to be what you are doing in your opening formula. But in Dirac notation you should not confuse matters by putting operators inside bra symbols or ket symbols. Inside the bra or ket should be simply a label: a symbol to say which bra or ket it is. The operator then acts on the whole thing. That is, $\hat{Q}$ acts on the ket $| \psi \rangle$ not on the label $\psi$.

You may be unfamiliar with writing an operator on the right of a bra, or as the last part of an expression. But it is in fact entirely logical and maps correctly to the mathematical behaviour of these mathematical objects. If in doubt, consider the case where an operator can be represented by a matrix and a ket can be represented by a vector.

[Final remark to pre-empt comments disagreeing with the above. Mathematical notation does not have to be Dirac notation, so if you have seen something like $| \hat{Q} \psi \rangle$ or $\langle \hat{Q} \psi|$ in a reputable text, it does not mean I am saying that text is wrong, I am just saying that at this point it departs from Dirac notation, and I would not recommend someone learning the subject to depart from well-formed Dirac notation.]

Further note

A much-used fact about kets and their decomposition is $$ | \psi \rangle = \sum_n | n \rangle \langle n | \psi \rangle $$ where $\{ |n \rangle\}$ is a complete orthonormal set of kets (also called state-vectors). By repeatedly using this one finds things like $$ \langle \phi | \hat{Q} | \psi \rangle = \sum_n \sum_m \langle \phi | n \rangle \langle n | \hat{Q} | m \rangle \langle m | \psi \rangle . $$ It follows that if $v_\phi$ is the vector whose components are $\langle n | \phi \rangle$ and $v_\psi$ is the vector whose components are $\langle n | \psi \rangle$ and $Q$ is the matrix whose components are $\langle n | \hat{Q} | m \rangle$ then $$ \langle \phi | \hat{Q} | \psi \rangle = v_\phi^\dagger Q v_\psi $$ where the expression on the right is written in the standard way for products of vectors and matrices.

For any such vectors and matrix it is the case that $$ v_\phi^\dagger Q v_\psi = (Q^\dagger v_\phi)^\dagger v_\psi $$ and I expect this explains why someone was proposing to move the $Q$ operator in front of the $\phi$ and put a dagger on it in the incorrectly-formed expression which the question asked about.

A similar fact arises for continuous variables.

In the case of a continuous set of states the sums become integrals, and if $| x \rangle$ is a position eigenstate then the quantity $\langle x | \psi \rangle$ is commonly written $\psi(x)$ and called a wavefunction. Here the letter $\psi$ is serving first as a label on a ket, and then as the name of a function. In this case the above expression becomes, for example, $$ \langle \phi | \hat{Q} | \psi \rangle = \int_{-\infty}^\infty \phi^*(x) Q \psi(x) dx $$ where $Q$ is now an operator acting on functions of $x$. And we can, if we wish, now use $$ \langle \phi | \hat{Q} | \psi \rangle = \int_{-\infty}^\infty (Q^\dagger \phi(x))^\dagger \psi(x) dx. $$ One can thus regard an operator either as acting to the right or as its adjoint acting to the left.

  • $\begingroup$ So, although $\langle\hat{Q}^\dagger\phi|$ is abuse of Dirac notation, could I take that as a meaning $\langle\hat{Q}^\dagger\phi|$ is bra corresponding to arbitrarily defined as the label named ket $\left|\phi\right\rangle$ acted by $\hat{Q}$? $\endgroup$
    – XX X
    Mar 10 at 2:11
  • 2
    $\begingroup$ I added a section which I hope clears this up. $\endgroup$ Mar 10 at 9:39

$$ \langle\varphi |\hat{O} = \langle{\hat{O}}^\dagger\varphi | .$$ Thus, $$ \langle\varphi |\hat{O} =\left [\begin{matrix}1\\1\\\end{matrix} \right ]^\dagger\left[\begin{matrix}1&i\\i&2\\\end{matrix}\right] =\left[\begin{matrix}1&1\\\end{matrix}\right]\left[\begin{matrix}1&i\\i&2\\\end{matrix}\right] =\left[\begin{matrix}1+i&2+i\\\end{matrix}\right]= \left[\begin{matrix}1-i\\2-i\\\end{matrix}\right]^\dagger \\ \langle{\hat{O}}^\dagger\varphi | =\left(\left[\begin{matrix}1&-i\\-i&2\\\end{matrix}\right]\left[\begin{matrix}1\\1\\\end{matrix}\right]\right)^\dagger = \left[\begin{matrix}1-i\\ 2-i\\ \end{matrix}\right]^\dagger =\left[\begin{matrix}1+i&2+i\\\end{matrix}\right] .$$

Likewise, integrating by parts, $$\int_{-\infty}^{\infty}{f^*\left (\left( -\hbar\frac{d}{dx}+m\omega x\right)g\right ) ~dx}\\ = \int_{-\infty}^{\infty}{\left (\left( \hbar\frac{d}{dx}+m\omega x\right)f\right )^*g~dx} ~~~.$$

So you might think of the gradient as an infinite-dimensional real antisymmetric matrix; and so, of the momentum as an imaginary hermitian one.

Dirac's notation merges the matrix and wave pictures: for $\hat D = -\hat D^\dagger = \int\!\! dx ~ |x\rangle \frac{d}{dx}\langle x|$ you see that $$ \langle \varphi |\hat D | \psi\rangle = \int\!\! dx ~ \langle \varphi|x\rangle \frac{d}{dx} \langle x|\psi\rangle \\ \equiv \int\!\! dx ~ \varphi(x)^* \frac{d}{dx} \psi(x)= \int\!\! dx ~\left (-\frac{d}{dx} \varphi(x) \right )^* \psi(x) \\ =\int\!\! dx ~ \langle \hat D^\dagger \varphi|x\rangle \langle x|\psi\rangle = \langle \hat D^\dagger \varphi| \psi\rangle . $$

  • $\begingroup$ I had already tried integrating parts as the question says, got $\int_{-\infty}^{\infty}{\left(\pm\hbar\frac{d}{dx}+m\omega x\right)f^*gdx}$. Of cource, maybe there are some wrongs, but I can't find. $\endgroup$
    – XX X
    Mar 10 at 1:24
  • $\begingroup$ I think carefully that $\int_{-\infty}^{\infty}{\left (\left( \hbar\frac{d}{dx}+m\omega x\right)f\right )^\dagger g~dx}$. $f^\dagger = f^*$ but $\left( \hbar\frac{d}{dx}+m\omega x\right)^\dagger \neq \left( \hbar\frac{d}{dx}+m\omega x\right)$ $\endgroup$
    – XX X
    Mar 10 at 1:48
  • $\begingroup$ It may be a waste of time for you to try to make me understand... Transposition of the derivative means integration parts? Why? I've never heard of it at all. Is it from the undergraduate course in physics? The gap between you and my knowledge is so wide that I don't seem to understand what you take for granted. Maybe it's better to just throw me a book that I can learn. I'm sorry that I'm stupid... $\endgroup$
    – XX X
    Mar 10 at 3:14
  • 1
    $\begingroup$ The book by Sakurai and Napolitano is quite clear… $\endgroup$ Mar 10 at 3:34

The following is a reply to the comment of XX X,

"It may be a waste of time for you to try to make me understand... Transposition of the derivative means integration parts? Why? "

We are not transposing the derivative operator. We are looking for its adjoint. Only in the case of a simple finite matrix representation do we obtain the adjoint by transposing and conjugation. The derivative operator is a bit more complicated than that.

The adjoint of an operator is defined with respect to an inner product. In the case of functions, we have the integral as inner product. The definition of the adjoint operator is given by $$ \langle \phi|\hat D|\psi\rangle =\langle \hat D^\dagger\phi|\psi\rangle\\ $$ Lets now use the position representation of the derivative operator $\hat D$ to work it out, $$ \langle \phi|\hat D|\psi\rangle =\int \phi^*(x)\frac{d}{dx}\psi(x)dx $$ We know that there is an adjoint operator corresponding to $\hat D$ that acts on $\phi$ without changing the value of the inner product, the integral. It turns out that the adjoint of the derivative operator in position representation is $-\frac{d}{dx}$ and not simply $\frac{d}{dx}$. You can proof that using integration by parts,

$$ \langle \phi|\hat D|\psi\rangle =\int \phi^*(x)\frac{d}{dx}\psi(x)dx = \int\left(-\frac{d}{dx} \phi^*(x)\right) \psi(x)dx=\langle \hat D ^\dagger\phi|\psi\rangle $$ By comparison you can conclude that the adjoint operator of $\frac{d}{dx}$ is $-\frac{d}{dx}$. This is sometimes written as $\left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx}$, or in the case of more than one variable, $\nabla^\dagger =-\nabla$. This can also be used to show that the momemtum operator is hermitian, $\hat p^\dagger = (-\hbar i \frac{d}{dx})^\dagger = -\hbar (i^\dagger) (\frac{d}{dx})^\dagger = -\hbar (-i)(-\frac{d}{dx})=-\hbar i\frac{d}{dx}=\hat p$.

I highly recommend reading the article "Mathematical surprises and Dirac's formalism in quantum mechanics" by F. Gieres, that can be found here https://arxiv.org/abs/quant-ph/9907069

It addresses the pitfalls that come with Dirac notation.

Answer to comment:

This line is incorrect, $$\langle \hat D ^\dagger\phi|\psi\rangle \neq \int_{-\infty}^\infty (D^\dagger \phi(x))^\dagger \psi(x) dx$$ The definition of the inner product expressed as integral is the following $$ \langle \hat D ^\dagger\phi|\psi\rangle = \int_{-\infty}^\infty (D^\dagger \phi(x))^* \psi(x) dx$$

Notice that the Bra only gets conjugated when you write down the inner product, it does not get adjungated a second time.

  • $\begingroup$ Thanks to your answer. Now I understand $$ \int \phi^*(x)\frac{d}{dx}\psi(x)dx = \int\left(-\frac{d}{dx} \phi^*(x)\right) \psi(x)dx $$ But the thing I've confused is why $$ \int\left(-\frac{d}{dx} \phi^*(x)\right) \psi(x)dx=\langle \hat D ^\dagger\phi|\psi\rangle $$ As far as I know $$ \langle \hat D ^\dagger\phi|\psi\rangle = \int_{-\infty}^\infty (D^\dagger \phi(x))^\dagger \psi(x) dx. $$ Isn't this the right way? $$ \int\left(-\frac{d}{dx} \phi(x)\right)^\dagger \psi(x)dx=\langle \hat D ^\dagger\phi|\psi\rangle $$ $\endgroup$
    – XX X
    Mar 10 at 12:37
  • 1
    $\begingroup$ @XXX I have updated my answer with regard to your comment. $\endgroup$
    – Hans Wurst
    Mar 10 at 13:55
  • $\begingroup$ Thank you very much! I've finally understood. In an integral, it is as if it had already been transposed. I did't have to do hermitian conjugate within the integral, I just have to do complex conjugate. $\endgroup$
    – XX X
    Mar 11 at 6:06

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