The following is a quote from an answer I was given to this previous question of mine:

The definition of the expectation value of an operator $A$ is $$\begin{equation} \langle A\,\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation}\tag{1}$$ (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position operator $$\begin{equation} \langle x\rangle=\int{x \psi^* (x) \psi (x) dx} \end{equation}\tag{2}$$

Ever since I asked that question I have been using equation $(1)$ without fully understanding why for a arbitrary operator $A$ its expectation value takes the form \begin{equation} \int{\psi^* (x) A(x) \psi (x) dx} \end{equation}

I understand that the probability is given by $$\begin{equation} \int{\psi^* (x) \psi (x) dx} =\int |\psi(x)|^2dx\end{equation}$$

So equation $(2)$ makes sense to me as it is simply $$\langle x \rangle=\int x|\psi(x)|^2dx$$ which is "the value of the variable times the probability of being in that configuration" as mentioned in the quote.

But for equation $(1)$ the arbitrary operator $A$ is $\color{red}{\text{in-between}}$ the $\psi^*(x)$ and $\psi(x)$.

So unless I can rewrite $(1)$ in the form of "value times probability": $$\begin{equation} \langle A\,\rangle=\int{A(x)\psi^* (x) \psi (x) dx}=\int A|\psi(x)|^2dx \end{equation}\tag{3}$$ I fail to see how equation $(1)$ gives the expectation value.

But I already know that equation $(3)$ is wrong since in equation $(1)$ the operator $A$ is acting on $\psi(x)$, so it doesn't make any sense to move the operator to the front of the integrand just to make it look like equation $(2)$.

Can anyone please explain to me why equation $(1)$ is justified as the expectation value even though the operator is in the middle?

  • $\begingroup$ What sort of explanation are you looking for here? The ultimate reason we use something that looks like $\vec v^T~\mathbf A\vec v$ is because it gives us linear algebra on our observables, which is highly desirable, but also allows those observables to be matrices which might not commute with each other, which gives us the uncertainty principle. $\endgroup$
    – CR Drost
    Dec 17, 2016 at 18:02

4 Answers 4


There is a slight, but important aspect you are missing here. The expectation value of the observable $A$ is defined as

$$\langle A\rangle_\psi=\int\psi^*(x)A\psi(x)dx$$

where as the probability of being in the configuration $\psi(x)$ is


But I already know that equation $(3)$ is wrong since in equation $(1)$ the operator $A$ is acting on $\psi(x)$, so it doesn't make any sense to move the operator to the front of the integrand just to make it look like equation $(2)$.

Yes of course. You are right. Now, we see the part you are missing. In quantum mechanics, we define the operators representing observables as Hermitian and an operator has got certain eigen functions.

If $\psi(x)$ is such an eigen function of the operator $A$, then you will have the eigen value equation


where $a$ is the corresponding eigen value which is a real number. In such a case,

$$\langle A\rangle_\psi=\int\psi^*(x)a\psi(x)dx=a\int\psi^*(x)\psi(x)dx=aP$$

where $P$ as defined above is the probability that the system can be found in the state $\psi(x)$. Hence we can say that the expectation value of an operator w.r.t a particular state is the eigen value of that state times the probability of being in that state. That's the difference between an expectation value and the eigen value.

Unless the wavefunction is normalized ($P=1$), we will not get the eigen value of the operator as it's expectation value.

Now, the wavefunction $\psi(x)$ need not be always an eigen function of $A$. In such cases, we expand our wavefunction as a superposition of the eigen functions of the operator $A$ in Dirac's bra-ket notation:

$$\vert \psi\rangle=\int d\zeta '\vert\zeta'\rangle\langle\zeta'\vert\psi\rangle $$

where {$\zeta_j$} forms a complete set of eigen functions of $A$ and $\displaystyle{\int d\zeta '\vert\zeta'\rangle\langle\zeta\vert}$ is the identity operator $1$ and $\vert\zeta'\rangle\langle\zeta'\vert$ is the projection operator $\Lambda_{\zeta'}$. The operations all happen in the appropriate Hilbert space spanned by the complete inner products of the eigen kets and eigen bras of the operator.

Before we proceed further, let's have a short brief on Dirac's formalism:

Short brief on Dirac's bra-ket notation: The ket, like the wavefunction represent a particular state of the system, but it's not actually the wavefunction of the system. It is represented as $\vert\psi\rangle$. The wave function of the system can be derived from the ket, and the ket representing a state, called the state ket, is a vector in the vector space spanned by the eigen kets of the operator $A$, just as like we speak the eigen functions of the operator $A$. Now, for the wavefunction, we have a corresponding complex wave function. Similarly, the complex dual of a state ket is called a state bra and is represented by $\langle\psi\vert$.
So, expectation value of some operator of the quantum mechanical system is what we want to measure. The first thing we consider is that we represent the general state ket (which is of course undefined) as a linear superposition of the eigen kets of the operator (which are known, once you solve the eigen value equation). It's like writing a vector as a linear combination of the independent coordinates. However, a vector space is a different thing. But the concept is the same. So, a general state ket $\vert\alpha\rangle$ can be expanded in terms of the complete eigen vectors of the operator $A$ as: $$\vert\alpha\rangle=\sum_{a'}c_{a'}\vert a'\rangle=c_{a'}\vert a'\rangle+c_{a''}\vert a''\rangle+c_{a'''}\vert a'''\rangle+...$$ where the kets $\vert a'\rangle, \vert a''\rangle,\vert a'''\rangle...$ are the eigen kets of $A$ and are complete. The set {$a'$} are the corresponding eigen values. The expansion coefficients $c_{a'},c_{a''},...$ are the probability amplitudes of the corresponding eigen kets. This can be understood in the coming paragraphs where we define the inner product of a ket and a bra.
We represent the state of the system in question as a linear combination of the eigen kets of the observable, whose expectation value is to be measured. This vector is represented as a ket and is defined in a complex vector space called the ket space. So, the ket space is spanned by the eigen kets of the operator. This means the eigen kets of the operator forms the basis vectors of our vector space. Since there is a one-to-one correspondence between a ket and the corresponding bra, we can define a space spanned by eigen bras and is called a bra space. If we take the inner product of the state ket and the state bra, defined respectively in the ket space and the bra space, we will get a complete inner product space called the Hilbert space. All the quantum "mechanics" happen in the Hilbert space.
Why do we need an inner product space? Well, the ket and bra are complex vectors and they are useless, unless we can extract some information from them. To obtain that, we take the inner product of the ket and bra. The inner product is taken between a bra and a ket. The inner product between the state ket $\vert\alpha\rangle$ and the state bra $\langle\beta\vert$ is denoted as $\langle\beta\vert\alpha\rangle$. It gives the probability amplitude that the system, found initially in the state $\vert\alpha\rangle$ to be found in the state $\vert\beta\rangle$, whose square of the modulus gives the probability of the same. The inner product is a real number. This probability is the fundamental thing that accompanies all the rest of the operations, which you will see in the coming discussions. The probability is a real number and must be positive. So the inner product explained above should be positive.
Now lets look back where we defined $c_{a'}$ as the probability amplitude of the state defined by the ket $\vert\alpha\rangle$ to be found in the state $\vert a'\rangle$, which is an eigen state of the operator $A$. For that, we take the inner product of $\vert\alpha\rangle$ with the eigen bra $\langle a'\vert$, we get
$$\langle a'\vert\alpha\rangle=\sum_{a'}c_{a'}\langle a'\vert a'\rangle=c_{a'}$$ where we have used an important relation called the orthonormality condition of two kets. If two kets $\vert a'\rangle$ and $\vert a''\rangle$ are orthogonal (independent) and normalized (so that the inner product of the ket with it's own bra gives $1$), then the orthonormality condition states that $$\langle a'\vert a''\rangle=\delta_{a',a''}$$ which is $1$ if the two kets are the same and $0$ when they are not. So, we demand the eigen kets of the operators to be orthonormal so that they satisfy the above orthonormality condition. So, we have got $c_{a'}$ as the probability amplitude of the eigen ket $\vert a'\rangle$. Hence the square of its modulus give us the probability that the system is found to be in the eigen state $\vert a'\rangle$: $$\vert c_{a'}\vert^2=\vert\langle a'\vert\alpha\rangle\vert^2$$ Now, we see that $$\sum_{a'} \vert c_{a'}\vert^2=\sum_{a'}\vert\langle a'\vert\alpha\rangle\vert^2=1$$ a requirement by the probability conservation theorem.
Now, what happens if we take the inner product of a general ket and the corresponding bra? That answer will give us the probability to find the system to be in that state. If the state kets are normalized, then this probability will be one.
Now, while taking the inner product of a state ket with a state bra, we are combining the two spaces- the ket and bra spaces- somehow to get a complete inner product space called Hilbert space. All the information about the state is hidden in this Hilbert space. So we ask the state ket to reveal some information, for example the energy. We do this by operating the state ket buy the energy operator. Then we will get the value of energy, which is present in the Hilbert space. So, the operations on state ket happens in the Hilbert space.
Now, let's see the operation of the operators on the state kets. Its similar to the operation of the operators on a wavefunction. The operator $A$ acting on the general ket $\vert\alpha\rangle$ is given by
$$A\vert\alpha\rangle=A\sum_{a'}c_{a'}\vert a'\rangle=A\sum_{a'}\left(\langle a'\vert\alpha\rangle\right)\vert a'\rangle=A\sum_{a'}\vert a'\rangle\langle a'\vert\alpha\rangle$$
When we compare both sides that the effect of $\displaystyle{\sum_{a'}\vert a'\rangle\langle a'\vert}$ is just like operating by the identity operator $1$. Hence $\displaystyle{\sum_{a'}\vert a'\rangle\langle a'\vert}=1$ is regarded as the identity opertor. Now, what does the outer product $\Lambda_{a'}=\vert a'\rangle\langle a'\vert$ gives us? Even though the inner product is a scalar, the outer product is an operator. To see this, let it act on the ket $\vert\alpha\rangle$
$$\Lambda_{a'}\vert\alpha\rangle=\vert a'\rangle\langle a'\vert\vert\alpha\rangle=\vert a'\rangle\left(\langle a'\vert\vert\alpha\rangle\right)=c_{a'}\vert a'\rangle.$$
The ket $\vert\alpha\rangle$ is a combination of the all possible eigen kets. When we operate this ket with $\Lambda_{a'}$, the operator selects the portion of the ket $\vert\alpha\rangle$ parallel to $\vert a'\rangle$. Hence it is known as the projection operator. Comparing the identity operator and the projection operator, we find that
$$\sum_{a'} \Lambda_{A'}=1$$
Okay, now we are almost equipped with the tools for the further discussion. We have only considered above discrete spectrum cases only. The above facts holds for continuous spectrum. All we have to do is just replace the summation by an integral and the Kronecker delta symbol by the Dirac delta function.
Note: This is not a complete description about Dirac's notation. There are a lot of things to see. However I've limitations here. You can found more illuminating discussions on Dirac's notation in Modern Quantum Mechanics by J. J. Sakurai.

Now, we continue. The expectation value is defined as

$$\langle A\rangle_\psi=\langle\psi\vert A\vert\psi\rangle$$

Substituting the above expansion of $\vert\psi\rangle$ in the equation, we get

$$ \begin{align} \langle A\rangle_\psi&=\iint d\zeta'd\zeta''\langle\psi\vert\zeta'\rangle\langle\zeta'\vert A \vert\zeta''\rangle\langle\zeta''\vert\psi\rangle\\ &= \iint d\zeta'd\zeta''\langle\psi\vert\zeta'\rangle\zeta' \delta\left(\zeta''-\zeta'\right)\langle\zeta''\vert\psi\rangle\\ &=\int d\zeta' \zeta' \langle\psi\vert\zeta'\rangle\langle\zeta'\vert\psi\rangle \end{align} $$

Now, $\langle\zeta'\vert\psi\rangle$ is defined as an inner product of two kets. It gives the probability that the system is transferred from state $\vert\psi\rangle$ to the state $\vert\zeta'\rangle$ and is the transition probability. If I represent $\langle\zeta'\vert\psi\rangle=c_{\zeta'}$, which in general is a complex number and is the transition amplitude, then $\langle\psi\vert\zeta'\rangle=\langle\zeta'\vert\psi\rangle^*=c^*_{\zeta'}$. Hence

$$\langle A\rangle_\psi=\int d\zeta ' \zeta' \vert c_{\zeta'}\vert^2$$

which means the expectation value ofthe operator $A$ is the eigen ket of $A$ times the probability of the system to be found in that particular eigen state of $A$.

  • 2
    $\begingroup$ +1--I like your way of phrasing it in terms of the eigenfunctions of $A$ (that's how I think about it), but I'm doubly impressed by the sidebar introducing Dirac notation :) $\endgroup$
    – Andrew
    Dec 17, 2016 at 23:01
  • 1
    $\begingroup$ @Unnikrishnan Thank you very much for a truly excellent answer. There were lots of good answers to this question, but in my opinion this is the best one. To me, it was this part of your answer that was the most vital: "If $\psi(x)$ is such an eigenfunction of the operator $A$, then you will have the eigenvalue equation $A\psi(x)=a\psi(x)$ where $a$ is the corresponding eigenvalue which is a real number. In such a case, $\langle A\rangle_\psi=\int\psi^*(x)a\psi(x)dx=a\int\psi^*(x)\psi(x)dx=aP$ where $P$ as defined above is the probability that the system can be found in the state $\psi(x)$". $\endgroup$
    – BLAZE
    Dec 18, 2016 at 7:05
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    $\begingroup$ @BLAZE that's a special case. In general one couldn't demand that the state ket in question should be an eigen ket of the operator. In that case, we expand it in terms of the eigen kets of the operator. I have mentioned that in the next paragraph. That's what the second part of my answer is all about. $\endgroup$
    – UKH
    Dec 18, 2016 at 17:33
  • $\begingroup$ @Unnikrishnan Yes, I noted that was not the general case but it just helped explain the form of the expectation value. In the QM course I am taking we were told that $\psi(x)=\sum\limits_n{a_n}\phi_n(x)$, is this analogous to the $\vert \psi\rangle=\int d\zeta '\vert\zeta'\rangle\langle\zeta'\vert\psi\rangle$ that you wrote? My lecturer doesn't use the Dirac notation. So it makes it hard for me to understand it. In any case the answer you gave here provides an explanation for it and will serve as very good guide for referring to in the future, thanks. $\endgroup$
    – BLAZE
    Dec 18, 2016 at 19:06
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    $\begingroup$ @PhyCoMath Its not an error. I meant complex dual itself. Its like an image and its mirror image. You have a space of images, which we call kets and the complex dual corresponds to the space of mirror images, where each element is called a bra. For each image in the image space, there is a corresponding image in the mirror image space. This mapping is what do we mean by a duality. The kets and bras are abstract mathematical objects (which we call vectors) that are elements of a vector space called the Hilbert space. $\endgroup$
    – UKH
    Dec 2, 2018 at 11:36

An expectation value is a probability weighted average.

Recall that the eigenvalues of an operator $\hat{Q}$ are the possible results of a measurement of the related observable $q$, and that $$ \hat{Q}\left|\psi_i\right\rangle = q_i \left|\psi_i\right\rangle \;,$$ where the $\psi_i$s are the eigenstates of $\hat{Q}$. From this it follows1 that given a state written in the basis of $\hat{Q}$ $$ | \psi \rangle = \sum_i c_n |\phi_i\rangle \;,$$ we can write the expectation value as \begin{align*} \left\langle\psi\right| \hat{Q}\left|\psi\right\rangle &= \sum_i \left\langle\psi_i\right|c^*_i \hat{Q} c_i \left|\psi_i\right\rangle \\ &= \sum_i c_i^* c_i \left\langle\psi_i\right| q_i \left|\psi_i\right\rangle \\ &= \sum_i c_i^* c_i \left\langle\psi_i|\psi_i\right\rangle\\ \tag{1} &= \sum_i P(q_i) q_i \;, \end{align*} where $P(q_i) = c_i^* c_i$ is the Born-rule probability of finding $q_i$ in your measurement.

But because the normalization requirement for $\left|\psi\right\rangle$ is $\left\langle\psi|\psi\right\rangle = \sum_i \langle \psi_i|c_i^* c_i | \psi_i \rangle = \sum_i P(q_i) = 1$, equation (1) is exactly the definition of a weighted average of $q$ over the probabilities $P(q_i)$.

1I'm going to do this in the form for a operator with a discrete spectrum, but the transition to a continuous space follows in the usual way.

  • 1
    $\begingroup$ +1, very elegant and gets to the point quickly. Just a thought, it might be worth stating explicitly that $|\psi_i\rangle$ is an eigenstate of $Q$, and not an arbitrary state (although maybe it's obvious from context). $\endgroup$
    – Andrew
    Dec 17, 2016 at 23:03
  • 1
    $\begingroup$ @Andrew That's a good point. I've just been teaching this same material and that detail went into the patter rather than the stuff actually written on the board, so I forgot to be explicit here. $\endgroup$ Dec 17, 2016 at 23:08

This is essentially the Born interpretation of the wave function.

Let $A$ be an observable. For the purpose of illustration, let me limit myself to the following simple scenario:

  1. There exists an orthonormal set of eigenfunctions $u_n(x)$ of $A$. This means that $$Au_n=a_nu_n,\\\intop \text d x u_n(x)^*u_m(x)=\delta _{nm},\\\psi =\sum _n c_n u_n. $$ The last equation is assumed to be valid, with appropriate coefficients, for every wave functions $\psi$. The coefficients may be found from the second equation, if the functions $u_n$ are known.
  2. All eigenvalues are non degenerate: $a_n=a_m \implies n=m$.

Now, to your question. Born's interpretation$^1$ tells us to compute the coefficients $c_n$ of the wave function $\psi$ and to take the squares of the moduli $\vert c_n\vert^2$. If we perform an experiment which is able to determine with certainty the value of the observable $A$, the probability to find $a_n$ is $\vert c_n\vert^2.$

That's all, this is how quantum mechanics works: it tells (1) which are the possible outcomes of the measurement of $A$ - just compute the eigenvalues of $A$ - and (2) what is the probability of a certain outcome - just compute the coefficients of $\psi$.

Now, the expectation value rule is just a corollary of Born's interpretation, since any expectation value should be defined by:$$\langle A \rangle = \sum _n a_n p_n,$$ where $p_n$ is the probability of the value $a_n$. If you plug $p_n=\vert c_n\vert ^2$ and use the eigenvalue equation satisfied by every $u_n$, you will discover that this definition is equivalent to: $$\langle A\rangle = \intop \text d x \psi^* A\psi.$$

$^1$ I don't know the history of quantum mechanics. No idea if this is how Born would have spelled his interpretation, probably not. This is what you will find in modern textbooks, and it is (more or less) the way physicists use $\psi$ to compute things.


By definition, $$ \langle A\rangle_\psi\equiv\langle\psi|A|\psi\rangle\tag{1} $$

Now plug $$ 1=\int\mathrm dx\ |x\rangle\langle x|\tag{2} $$ into the definition of $\langle A\rangle_\psi$, and use $$ \begin{aligned} \psi(x)&\equiv\langle x|\psi\rangle\\ \langle x|A|x'\rangle&\equiv A(x)\delta(x-x') \end{aligned}\tag{3} $$

More details: $$ \begin{aligned} \langle A\rangle_\psi&\overset{(1)}\equiv\langle\psi|A|\psi\rangle=\\ &\overset{(2)}=\langle\psi|\overbrace{\left[\int\mathrm dx\ |x\rangle\langle x|\right]}^1A\overbrace{\left[\int\mathrm dx'\ |x'\rangle\langle x'|\right]}^1|\psi\rangle\\ &=\int\mathrm dx\,\mathrm dx'\ \langle\psi|x\rangle \langle x|A|x'\rangle \langle x'|\psi\rangle\\ &\overset{(3)}=\int\mathrm dx\,\mathrm dx'\ \psi^*(x)A(x)\delta(x-x')\psi(x')\\ &=\int\mathrm dx\ \psi^*(x)A(x)\psi(x) \end{aligned} $$

  • $\begingroup$ This is the most basic description one can give @BLAZE; you need to be more elaborate on what step you have stumbled down. Is it the Dirac Delta function? $\endgroup$
    – user36790
    Dec 17, 2016 at 11:38
  • $\begingroup$ @AccidentalFourierTransform That's looking better thanks, I am frantically searching the internet trying to find explanations for $1=\int\mathrm dx\ |x\rangle\langle x|$ and $ \langle x|A|x'\rangle \equiv A(x)\delta(x-x')$. I'm afraid I have simply no idea what equation $(2)$ or $(3)$ actually mean (I'm just a first year undergraduate). Why is the Dirac-delta measure being used here? Should I just accept that it is the definition of expectation value and forget trying to understand why it is that way? Or is there a link you could provide that proves it? $\endgroup$
    – BLAZE
    Dec 17, 2016 at 11:39
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    $\begingroup$ @BLAZE, Please check the Feynman Lectures Vol III; it is explicitly well described there. $\endgroup$
    – user36790
    Dec 17, 2016 at 11:43
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    $\begingroup$ @BLAZE I fear that your questions are way too broad: you are asking me to provide an introduction to QM. The best I can do is to refer you to the standard literature. Griffiths' Introduction to QM is a popular choice, but I learnt with Cohen&Tannoudji's book, so this is the one I'd recommend. As mentioned by MAFIA, Feynman's lectures provide a very intuitive meaning to these equations as well (though one should remark that Feynman's approach is slightly outdated nowadays, but it is a must read anyway). $\endgroup$ Dec 17, 2016 at 11:43
  • 1
    $\begingroup$ And here is the link to the CalTech site. $\endgroup$
    – user36790
    Dec 17, 2016 at 11:45

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