For a while, I had difficulty understanding the differences between an expectation value and the state-vector derived from the Schrodinger equation. My understanding is that the Schrodinger equation returns a state-vector $\mid\Psi(t)\rangle$ that can be used to calculate probabilities for events corresponding to possible values, or eigenvalues, of a system. On the other hand, expectation values allow one to determine, with varying accuracy, what the value of an observable is 'most likely' to be, regardless of whether it is a possible state. For example, take a single spin. If the spin is prepared in some direction and rotated in an arbitrary direction, the spin will be either +1 or -1. However, the expectation value will be $\sigma_n = \langle 0 \rangle$. Solving the Schrodinger equation will give a ket $\mid\Psi(t)\rangle$ that can be used to calculate the probabilities that the spin will be +1 or -1.

Both of these are helpful tools used to find what the value of an observable may or may not be. Is there a mathematical or physical relationship between the two? Or are they just two separate ways of obtaining similar but fundamentally different information?

  • $\begingroup$ Are you wanting to know the difference between the solution to the Schrodinger equation, the Schrodinger equation itself or the state vector compared to the expectation value? It's not clear to me entirely because the title says one thing and the first sentence says another $\endgroup$
    – Andrew
    Apr 13 '20 at 9:31

$|\Psi(t)\rangle$ contains the time evolution of a superposition of states describing the probability amplitude of each state of the system - it is not a physical entity but an abstract vector representation of the probability amplitude of measurable states (energy, spin, position etc).

Let's ignore the time dependence of $\Psi$ for now since it's not needed to answer your question.

Suppose the system is in a superposition of observable states $A$, $|\Psi \rangle = A_1|A_1\rangle + A_2|A_2\rangle \:+ \; ...$

$A_i$ is the probability amplitude of state $|A_i\rangle$ where $i$ may be $1,2,3...$

The expectation value of $A$ is:

$|$average probability amplitude of all the superposition states of $A$ that $\Psi$ is in$|^2$

and can be calculated using the matrix representation of the operator representing measurable $A$:

$A \;\dot{=} \;\hat{A}$ (matrix form)

$\langle A\rangle\, = \,\langle\Psi|\hat{A}|\Psi\rangle =\sum\limits_{i}$$A_{i}P_{A_{i}}$

Where $P_{A_{i}}$ $= |\langle A_i|\Psi\rangle|^2$ which is the |probability amplitude of specific state $|A_i\rangle|^2$

Two conclusions to answer your question:

$\bullet$$\;\;$$\boldsymbol{|\Psi\rangle}$ is a superposition which means it represents the system being in a combination of "$A$" states simultaneously until we measure the state of the system (causing the collapse of $|\Psi\rangle$ to just one state of $A$ let's say is $|A_i\rangle$).

$\bullet$$\;\;$$\boldsymbol{\langle A\rangle}$ is the mean value of a large number of experiments measuring A. It is not a time average, but an average over many identical experiments.


The expectaion value of quantity $A$ is given by $$\langle A\rangle = \langle\Psi|\hat{A}|\Psi\rangle,$$ where $\hat{A}$ is the operator for this quantity, whereas $|\Psi\rangle$ is a state vector, also called a wave function.

As the system evolves in time, the expectation values also change. In the so-called Schrödinger picture/representation this is accounted for by evolving the state vectors according to the Schrödinger equation: $$i\hbar\partial_t|\Psi(t)\rangle = \hat{H} |\Psi(t)\rangle,$$ so that the time-dependent expectation values are given by $$\langle A(t)\rangle = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle.$$

Extra material
An alternative view is the so-called Heisenberg picture/representation, where the evolution is contained in the operators, which obey the Heisenberg equations $$\partial_t \hat{A}(t) = \frac{1}{i\hbar}[\hat{A},\hat{H}],$$ where $[...,...]$ designates a commutator and the expectation value is given by $$\langle A(t)\rangle = \langle\Psi|\hat{A}(t)|\Psi\rangle.$$

  • $\begingroup$ I wouldn't say the state vector is also called the wave function. It might lead to confusion. If we are trying to describe their relationship, it's more rigorous to say the wave function is a state vector projected onto, say, the x basis in the case that the wave function is a function of space. One is a vector and one is an algebraic expression $\endgroup$
    – Andrew
    Apr 13 '20 at 10:29
  • $\begingroup$ I agree that state vector is the correct mathematical term, whereas wave function is better reserved for $\langle x|\Psi\rangle$. However people often do use them interchangeably. $\endgroup$ Apr 13 '20 at 11:29

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