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Expectation value of any operator $\hat{Q}$ is defined as, $$ \left\langle\psi_n\mid\hat{Q}\mid \psi_n\right\rangle $$

and action of the operator $\hat{Q}$ on wavefunction is defined as $$ \hat{Q} \mid\psi_n\rangle= q \mid\psi_n\rangle $$

It is my understanding, that when we apply an operator (say $\hat{H}$) to a wavefunction it returns its eigenvalue corresponding to the wavefunction, which would be $E_n$ in this case, but isn't that just the same as $\left\langle\psi_n\mid\hat{H}\mid \psi_n\right\rangle $ ?

I'd like to know what the difference is between $\hat{Q}\mid\psi_n\rangle$ and $\left\langle\psi_n\mid\hat{Q}\mid \psi_n\right\rangle$, what exactly are you doing when you apply an operator to a wavefunction?

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    $\begingroup$ For eigenstates, the expectation is the eigenvalue. But for superpositions of eigenstates, the expectation weights eigenvalues with probabilities. $\endgroup$
    – J.G.
    Commented Apr 25 at 22:48
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    $\begingroup$ The expectation value is a number. The left and right sides of the eigenequation are functions. $\endgroup$
    – Ghoster
    Commented Apr 25 at 22:56
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    $\begingroup$ why are there close votes? This is kind of a tricky part of beginner QM, and sort of lead into the measurement problem which still generates lots of interpretation conversation. $\endgroup$
    – JEB
    Commented Apr 26 at 4:24
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    $\begingroup$ Do you know linear algebra? It might be worth to take a step back from the physics and just consider the math. Then the difference between both things should be clear. $\endgroup$ Commented Apr 26 at 7:29

5 Answers 5

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The expectation value of an observable $\hat{Q}$ is the average obtained by repeated measurements of $\hat{Q}$ (i.e., average of the eigenvalues of $\hat{Q}$). That is, it is the most likely outcome of a measurement of the observable $\hat Q$. Hence the word expectation (value).

But I do not understand how the expectation value relates to eigen function equations: $$ \hat{Q} \mid\psi_n\rangle= q \mid\psi_n\rangle $$

This equation states that a measurement of the physical observable $\hat Q$ gives us the value $q$ where $q$ is the physical quantity associated with $\hat Q$ (and of course $\mid\psi_n\rangle$ is an eigenstate of $\hat{Q}$).

There are cases where the eigenvalue and expectation value are the same. An example would be a quantum system where every outcome of the measurement of $\hat Q$ has the same probability.

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  • $\begingroup$ So the expectation value of an observable $\hat{Q}$, is the average of the values of $q$ that you would expect? This confuses me, as for a given $\hat{Q}$ and $\mid\psi_n\rangle$, there is only one $q$. $\endgroup$ Commented Apr 26 at 14:07
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Specifically, I'd like to know what the difference is between $\hat{Q} \psi_n$ and $\left\langle\psi_n\mid\hat{Q}\mid \psi_n\right\rangle$

Your $\hat Q|\psi_n\rangle $ represents an operator $\hat Q$ acting on a state $|\psi_n\rangle$, which will result in another state, e.g., $|\chi_n\rangle$ a la: $$ |\chi_n\rangle = \hat Q |\psi_n\rangle\;. $$

In the case where the state $|\psi_n\rangle$ is an eigenstate of the operator $\hat Q$, the state $|\chi_n\rangle$ is proportional to the state $|\psi_n\rangle$ and we can write: $$ \hat Q|\psi_n\rangle = q|\psi_n\rangle\;, $$ where $q$ is a number.


An expectation value has it's usual meaning from probability theory. Given a function $q(\vec x)$ and a probability density $p(\vec x)$, the expectation value is: $$ E_p[Q] = \int d x p(\vec x)q(\vec x)\;. $$

Or, if it is easier to think about in terms of discrete probabilities $p_i$, the expectation value is a sum: $$ E_p[Q] = \sum_i p_i q_i\;. $$


In quantum mechanics, the probability density is related to the state via $$ p_n(\vec x) = |\psi_n(\vec x)|^2\;, $$ where we are now thinking of the state $|\psi_n\rangle$ in a basis, such as for example the position $\vec x$ basis, such that $\psi_n(\vec x) = \langle \vec x | \psi_n\rangle$.

If an operator like $\hat Q$ is diagonal in the position basis, we can write: $$ \hat Q = \int d x |\vec x\rangle\langle\vec x |q(\vec x)\;, $$ where $q(\vec x)$ is a function (not an operator).

Then, we can write: $$ \langle \psi_n|\hat Q |\psi_n\rangle = \int d x \psi_n(\vec x)^* q(\vec x)\psi_n(\vec x) $$ $$ =\int d x |\psi_n(\vec x)|^2 q(\vec x) $$ $$ =\int d x p_n(\vec x) q(\vec x) $$ $$ =E_{p_n}[Q] $$


I do not understand how the expectation value relates to eigen function equations:

$$ \hat{Q} \mid\psi_n\rangle= q \mid\psi_n\rangle\;.\tag{A} $$

The above Eq. (A) is only true (by definition) when the state $|\psi_n\rangle$ is a eigenstate of the operator $\hat Q$. The eigenvalue is called $q$, and it is generally a complex number (or a real number if the operator is Hermitian).

If such a equation is true, this is the same as saying that the state has a definite value $q$ for the operator $\hat Q$. If the operator $\hat Q$ is a Hermitian "observable" (which thus has real eigenvalues and can correspond to real measurements) then we say that the measure value will be $q$ with 100% probability. This also means that the expected value (expectation value) will be $q$, which is trivial to see, since we assume the state is normalized: $$ \langle \psi_n|\hat Q|\psi_n\rangle = q\langle\psi_n|\psi_n\rangle =q\;, $$ where the first equality follow from Eq. (A).

However, in general, there is no reason to expect that Eq. (A) will hold for arbitrary states and operators. In a discrete basis, we would rather expect that generally: $$ \hat Q |\psi_n\rangle = \sum_{m}|\psi_m\rangle Q_{m,n}\;. \tag{B} $$

Eq. (A) is a special case of Eq. (B) when only one term in the sum contributes (the term with $m=n$, which we called $q=Q_{n,n}$).

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  • $\begingroup$ Thanks for your answer, my question has been edited, hopefully it is clearer $\endgroup$ Commented Apr 25 at 23:08
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The real difference results from arbitrary or mixed states.

$|\Psi\rangle=\sum c_k|\psi_k\rangle$

$\hat O|\psi_k\rangle=\lambda_k|\psi_k\rangle$

$1=\sum c_kc_k^*$

$\hat O |\Psi\rangle=\sum c_k \hat O|\psi_k\rangle=\sum c_k c_k^* \lambda_k|\psi_k\rangle$

$\langle \Psi|\hat O | \Psi \rangle=\sum c_kc_k^*\lambda_k=\langle \hat O \rangle$

The eigenvalue and the expectation value are the same for a single eigenstate. The relationship is more complicated for more than one state.

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Applying an operator to any vector, whether or not it is an eigenvector, returns a vector not an eigenvalue: $$X|\psi_n\rangle=x_n|\psi_n\rangle\neq x_n.$$

The expectation value of $X$ in the state $|\psi_n\rangle$ is a number and it happens to be the eigenvalue in this state: $$\langle\psi_n|X|\psi_n\rangle=x_n.$$

For any state $|\phi\rangle=\sum_n\alpha_nx_n$, we have: $$X|\phi\rangle=\sum_n\alpha_nx_n|\psi_n\rangle$$ which is a vector not a number and $$\langle\phi|X|\phi\rangle=\sum_n|\alpha_n|^2x_n$$ which is a number but may not be equal to any of the eigenvalues.

The product of a state and an operator gives you the result of performing some operation on the state. For example, the not operator applied to a qubit gives you the result of flipping that qubit.

The numbers $|\alpha_n|^2$ play the roles of probabilities in situations where the state has decohered, which includes many measurements:

https://arxiv.org/abs/2208.09019

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It could conceivably be argued that $\langle \psi | \hat{Q} | \psi \rangle$ has a physical meaning while $\hat{Q} | \psi \rangle$ does not (unless $\hat{Q}$ is a unitary or projector). The first is the average value that occurs when we measure $|\psi\rangle$ an infinite (or near infinite) number of times to determine the value of $Q$ (assuming $\hat{Q}$ is Hermitian). If $0\leq \hat{Q}^{\dagger} \hat{Q} \leq 1$ then we can interpret $\hat{Q}$ as an update and $\hat{Q}|\psi \rangle$ as a different (possibly unnormalised) quantum state. Otherwise, it is only a mathematical intermediary for determining something else. While I wrote this for general pure states $|\psi\rangle$, this is also true for eigenvectors of $\hat{Q}$, $|\psi_{n}\rangle$.

In contrast, $\hat{Q}|\psi_{n}\rangle = q_{n} |\psi_{n}\rangle$ has an interpretation as specifying an eigenvalue and eigenvector pair from $\hat{Q}$. This informs us that the quantity $Q$ is well defined for the state $|\psi_{n}\rangle$ and measurements of $Q$ will not change the state, which may not always be the case.

Or if you prefer a mathematical difference, $\langle \psi |\hat{Q}|\psi\rangle$ is a number, while $\hat{Q}|\psi\rangle$ is a vector. So they are used for different things.

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