# Difference between the expectation value of an operator and operator applied to wave function?

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?

• For eigenstates, the expectation is the eigenvalue. But for superpositions of eigenstates, the expectation weights eigenvalues with probabilities.
– J.G.
Commented Apr 25 at 22:48
• The expectation value is a number. The left and right sides of the eigenequation are functions. Commented Apr 25 at 22:56
• 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.
– JEB
Commented Apr 26 at 4:24
• 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. Commented Apr 26 at 7:29

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.

• 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 $q$. Commented Apr 26 at 14:07

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 $$q=Q_{n,n}$$).

• Thanks for your answer, my question has been edited, hopefully it is clearer Commented Apr 25 at 23:08

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.

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

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.