I've read wiki definition and also some other questions here on StackExchange, but I still can't quite catch what is a quantum state, what are properties of said state. I also was looking for some basic example of it but I couldn't find any. I would appreciate an explanation for a true beginner.
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1Could you elaborate on what exactly you don't understand? – Tobias Fünke Aug 31 '21 at 17:30
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1@Jakob well, thing that I don't understand is that, if it's a function, a lab set or something else or both. I don't know what associate it with. On wiki it is said that quantum state provides a probability distribution and I know what a probability distribution is (I think), but I don't understand what does it mean that quantum state "provides" it. I think a simple example of this state could help me understand it. – cover Aug 31 '21 at 17:39
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does a true beginner know what a classical trajectory is? Or perhaps, phase space? – JEB Aug 31 '21 at 19:10
3 Answers
State in general
Generally speaking, a state of a system is some mathematical object that completely describes the system at a particular time. For example, the state of a free classical particle is identified by its position $x$ and momentum $p$ and the state of a computer is identified by the values stored in all its registers and memory. In many theories, there are two fundamental calculations that involve a state. First, we can use the laws that govern state evolution to compute the state at time $t_2$ based on the state at time $t_1$. Second, we can use a state to predict the outcomes of observations.
This general picture applies to classical mechanics, thermodynamics, chemistry, control theory, software engineering, quantum mechanics etc.
State in quantum mechanics
Perhaps the best way to intuit what the quantum state is is to consider how it connects to observable quantities. In quantum mechanics, measurement is probabilistic and the probability distribution over outcomes of a measurement is constructed from two ingredients:
- an observable $A$, which is a mathematical object describing the measurement to be performed,
- a quantum state $\rho$, which is a mathematical object describing the quantum system at a given time$^1$.
Given these two ingredients one can compute the probability $p(\lambda|\rho, A)$ that the measurement of $A$ on a system in state $\rho$ yields the outcome $\lambda$. The recipe for computing $p(\lambda|\rho, A)$ is known as the Born rule.
Quantum theory also tells us how quantum states evolve, so we can find the state $\rho_2$ of our system at time $t_2$ from its state $\rho_1$ at time $t_1$. See Schrödinger equation for details.
Example
The simplest quantum mechanical system is a qubit. Observables on a qubit are modeled as complex $2\times 2$ Hermitian matrices and states are modeled as complex $2\times 2$ positive semidefinite matrices with unit trace. Thus, the following matrix is an example state of a qubit
$$ \rho = \frac12\begin{bmatrix}1&1\\1&1\end{bmatrix} = |+\rangle\langle +|. $$
We can also consider an example observable
$$ A = \begin{bmatrix}1&0\\0&-1\end{bmatrix} = |0\rangle\langle 0|-|1\rangle\langle 1| $$
to see how $\rho$ determines probabilities of the outcomes of a measurement of $A$. The eigenvalues of $A$ are $1$ and $-1$ and so the only possible outcomes of a measurement of $A$ are $1$ and $-1$. The Born rule says that
$$ p(\lambda|\rho,A) = \mathrm{tr}(\rho P_\lambda)\tag1 $$
where $P_\lambda$ is the projector onto the eigenspace of $A$ associated with $\lambda$. Let's use $(1)$ to compute the probability that our measurement of $A$ on a qubit in state $\rho$ yields $-1$. Noting that $P_{-1} = \begin{bmatrix}0&0\\0&1\end{bmatrix}$, we have
$$ p(-1|\rho,A) = \mathrm{tr}\left( \frac12 \begin{bmatrix} 1&1\\1& 1\end{bmatrix} \cdot \begin{bmatrix}0&0\\0&1\end{bmatrix} \right) = \frac12. $$
This exemplifies how for each observable $A$ our state $\rho$ determines the probability distribution over outcomes.
$^1$ We use the Schrödinger picture here.

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A nitpick: "a state of a system is some mathematical object that identifies the configuration of a system at a particular time." I think it's a bit confusing to use the word "configuration" because as you say, the state of a classical particle consists of both $x,p$ but a point in the configuration space of the particle consists of only $x$. – Aug 31 '21 at 19:25
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A more serious point tho, a density matrix is not the state of a quantum system. A state vector in the Hilbert space is a state of a quantum system. Sure, you can represent a quantum system that can be described by a state vector also via a density matrix but the converse is not true, you can write down a density matrix also for a quantum system that simply does not have a state: for example, for one of the qubits in a Bell pair. – Aug 31 '21 at 19:30
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Also, the first density matrix you wrote does not equal to $\vert +\rangle\langle +\vert$. To explicitly check this: If you square the LHS, you don't get the same object back, if you square the RHS, you do get the same object back. But more simply, a density matrix proportional to the identity matrix corresponds to a maximally mixed state so it cannot be equal to a density matrix constructed out of a pure state. Registering a $-1$ for now. – Aug 31 '21 at 19:33
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Re "configuration": I didn't mean "configuration" as in "a point in the configuration space", but I agree this type of potential for confusion should be avoided. Edited. – Adam Zalcman Aug 31 '21 at 19:37
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@DvijD.C. A vector on the Hilbert space is not a quantum state. A state is a linear functional. In finite dimensions, every state can be defined via $\omega(\cdot) = \mathrm {Tr} (\cdot \rho)$ for some density operator $\rho$ and vice versa: That is, every density operator defines a state. – Tobias Fünke Aug 31 '21 at 19:41
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2Re density matrix: I disagree. You appear to be using the term "state" to denote what most people call "pure state", implying that mixed states aren't states. This is not standard terminology. – Adam Zalcman Aug 31 '21 at 19:43
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@Jakob OK, I agree -- if one is to consider a mixed state also as a state. However, in my experience, when one says a state, it is taken to mean a pure state unless otherwise specified. However, that's neither here nor there so I retract my comment about states. But, I think the distinction ought to be made: for example, when one talks in terms of wavefunctions, it becomes abundantly clear that a mixed state does not have a wavefunction of its own, only the full system has a wavefunction. – Aug 31 '21 at 19:46
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Re my first density matrix: I disagree once more. The square of LHS equals LHS, as required. Also, LHS is not proportional to the identity matrix. – Adam Zalcman Aug 31 '21 at 19:47
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1@AdamZalcman Ah, my mistake, I misread your first density matrix. Retracting the vote. Thanks. – Aug 31 '21 at 19:48
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2Just an addendum to my previous comment: In the infinite-dimensional case, there are states which cannot be represented by a density operator, cf. this or this. – Tobias Fünke Aug 31 '21 at 19:49
There's a very good discussion of this in a paper with almost the same title: "Peres, Asher. "What is a state vector?." American Journal of Physics 52.7 (1984): 644-650."
Peres argues that a state vector (or quantum state) represents a procedure. In Sec. III, Peres quotes "Giles, R., 1970. Foundations for quantum mechanics. Journal of Mathematical Physics, 11(7), pp.2139-2160"
"We define a state of a system to be a (nonempty) collection of methods of preparation of the system, where by a method of preparation we mean a document giving detailed instructions for the preparation".
In other words, writing $\vert\psi\rangle$ should be understood as meaning that some manipulations have been made and we give an interpretation of result of these manipulations as the quantum state $\vert \psi\rangle$.
The discussion of Peres is quite approachable, and he is careful to give examples of contradictions with various other interpretations of a state. There's a follow up discussion on the time evolution of quantum states.

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Philosophically this is nice, but does Asher explain why/how to represent a "method of preparation" as a vector in a Hilbert space (that can be "scaled" and "added", an acted upon linearly by matrices)? – D.R Apr 30 '23 at 19:55
I think the best way to describe quantum states, is by realizing that they are just a mathematical way of representing real structures or phenomena. The real interpretation of the quantum wave function is still an open question, however, the one thing we do know is, different structures are represented through different states.
Imagine an electron in the hydrogen atom. Forget all the advanced quantum mechanics for a second. Imagine, the electrons are particles revolving in $K,L,M,N...$ shells. From Bohr's theory, you can easily check that electrons in different shells have different energies. The point is, you can have an hydrogen atom in a ground state, where the electron is in $K$ shell. However, you can also have an atom in an excited state $M$ shell, for example. Even though these are both hydrogen atoms, they are different. The quantum wavefunction or 'state' is a mathematical way of representing these different structures, so that you can easily differentiate between them.
Similarly, there are other examples of this. What I'm getting to, is that a quantum 'state' does not have different properties as such. It is a 'collection' of all the properties of the system you are describing. They are a way of mathematically representing a system, like the different hydrogen atoms in the previous example. It is a mathematical description of a system, that allows you to distinguish or compare it to another system.

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