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$|\psi\rangle$ is the quantum state of a system and $\psi(x)$ is the quantum wave function. Quite simply, what is the difference?

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    $\begingroup$ Do you understand the meaning of $\psi(x) = \langle x | \psi \rangle$? $\endgroup$ – DanielSank Nov 2 '20 at 17:29
  • $\begingroup$ no please explain $\endgroup$ – STEMguy24 Nov 2 '20 at 17:35
  • $\begingroup$ This is a good question which is often unhelpfully overlooked in introductory courses. Maybe also see here. $\endgroup$ – Charlie Nov 2 '20 at 18:04
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A quantum state $|\psi\rangle$ is an abstract vector in a vector space (usually called the state space) that describes the state of your system. This state space is a complex vector space, and usually has many (infinite) dimensions. However, one can still draw some analogies with the usual Euclidean 3D vector space. We are being somewhat lax when doing that, but it should hopefully help understanding.

In the 3D Euclidean space, you have an abstract vector $\mathbf{r}$ that for example encodes the position of a particle. This is equivalent to the abstract vector $|\psi\rangle$ in the state space that encodes the quantum state of the system. When you do practical calculations, you write abstract vectors in a basis. In the 3D Euclidean space, this could be:

$$ \mathbf{r}=a\hat{\mathbf{x}}+b\hat{\mathbf{y}}+c\hat{\mathbf{z}}, $$

where $(\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}})$ are the basis vectors. The components of $\mathbf{r}$ in this basis are given by the scalar products: $a=\hat{\mathbf{x}}\cdot\mathbf{r}$, $b=\hat{\mathbf{y}}\cdot\mathbf{r}$, $c=\hat{\mathbf{z}}\cdot\mathbf{r}$.

You can do the same in state space. You define a basis in which you can write your abstract state $|\psi\rangle$. One possible basis is the "position basis", given by the eigenstates of the position operator, usually labelled $\{|x\rangle\}$, where $x$ is a 1D coordinate (there are some subtleties about the position basis that I am not getting into here). You can also write the abstract vector $|\psi\rangle$ in this basis, and you get:

$$ |\psi\rangle=\int \text dx\,\psi(x)|x\rangle. $$

The basis states $|x\rangle$ here play the same role that the basis states $(\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}})$ play in the Euclidean space; the components $\psi(x)$ play the same role that the components $(a,b,c)$ play in the Euclidean space; and the sum over the three direction in the Euclidean space is here replaced by an integral over $x$ because the basis in this case is given by a continuous variable. Just like you can find the components of the vector $\mathbf{r}$ in the Euclidean case by calculating the scalar product with the basis vectors, you can do the same in the state space, and the corresponding expression is

$$ \psi(x)=\langle x|\psi\rangle. $$

Therefore, the wave function $\psi(x)$ gives the components of the abstract state $|\psi\rangle$ in the position basis.

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  • $\begingroup$ Sorry I'm not a genius. Can you explain this so i can understand it $\endgroup$ – STEMguy24 Nov 2 '20 at 18:04
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    $\begingroup$ @STEMguy24 Perhaps you could clarify what you do know so that we know at what level to pitch the answer. For example, do you understand from my answer what a 3D vector r is and what its components (a,b,c) are when you write it in a basis? $\endgroup$ – ProfM Nov 2 '20 at 18:46
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There are three related concepts here which are different and people often use them interchangeably and it is mostly clear in the context as to what is meant, however, keeping the distinction in mind is important so as to avoid possible conceptual confusion.

  • Quantum State: As one can imagine, this corresponds to the object in our formalism that is supposed to stand in one-to-one correspondence with the physical state of the system that we are describing. In other words, this is the formal object which tells us everything there is to know about the physical state of the system that we are interested in. In quantum mechanics, this means that this is the object that tells us the probabilities of obtaining various outcomes corresponding to all possible observables that can be measured for the given system.

    It turns out that complex vectors (of a unit norm) who live in a Hilbert space are the mathematical objects that are well-suited to do this job. However, there is one caveat: the overall phase of a complex vector that is used to describe the physical state of a quantum system is physically irrelevant and thus the formal object that stands in one-to-one correspondence with the physical state of a system should not contain the information about this overall phase factor. Thus, the formal object that uniquely encodes the physical state of a quantum system for us in the formalism of quantum mechanics is a ray in the (relevant) Hilbert space. A ray is a set of complex vectors in the Hilbert space that only differ from each other by a scalar factor of the unit norm (i.e., by an overall phase ).
  • State Vector: In continuation of the discussion in the above paragraph, a state vector is a particular complex vector belonging to a ray. It is often convenient to write down specific complex vectors to describe a physical system instead of referring to a collection of complex vectors who differ from each other by an overall phase. So, let's say a ray $\mathcal{R}$ corresponds to the physical state of a system $S$ then there would be many complex vectors in the Hilbert space that belong to this ray $\mathcal{R}$, such as $\vert\psi\rangle$ and $e^{i\alpha}\vert\psi\rangle$ (where $\alpha\in\mathbb{R}$). Each one of them is called a state vector representing the system $S$. In practice, we mostly use state vectors (with the implicit understanding that the overall phase is irrelevant) rather than using rays.
  • Quantum Wave Function: The quantum wave function is the expression of the state vector via the coefficients in the expression of the state vector as a linear combination of the vectors in a basis (a complete set of vectors). So, the idea is that you can convey the full information that is contained in a state vector $\vert\psi\rangle$ by giving the list of coefficients that arise when you express the state vector as a linear combination of the vectors in the said complete set of vectors. In quantum mechanics, when the complete set of vectors is chosen to be formed by the eigenvectors of some observable, these coefficients are of particular interest because they represent the probability amplitudes of the given physical state of the system over the said eigenvectors.

    For example, we can chose the eigenstates of the position operator to be the complete set of vectors for the Hilbert space of a spinless particle in one dimension. In this case, some state vector $\vert\psi\rangle$ can be represented as the collection of probability amplitudes $\psi(x)\equiv \langle x\vert\psi\rangle$. As you can see, this is the familiar face of the wave function. So, the quantum wave function is an even more "practical" way of describing the physical state of a quantum system because these probability amplitudes are pure numbers (albeit, complex).
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A quantum state is an abstract mathematical entity, denoted by $\left|\Psi\right>$. Its sole characteristic is that it is element of a Hilbert space: $\left|\Psi\right> \in \mathcal{H}$. We know $\left| \Psi \right>$ so far as we know the structures on $\mathcal{H}$.

One class of things that live in the environment of a Hilbert space is the class of operators.

Given a hermitian operator $N: \mathcal{H} \to \mathcal{H}$ --and turning a blind eye to the mathematical subtleties for the sake of the idea-- we can decompose the abstract $\left|\Psi\right>$ as the sum of the eigenvectors times coefficients $$\left| \Psi \right> = \sum_n \left| n \right>\left<n\right. \left.|\Psi\right>.$$

It is the coefficients of this decomposition, $\left<n\right. \left.|\Psi\right>$, that you call wave function.

The same abstract $\left| \Psi \right>$ can have many different representations.

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Remember that - because our observables correspond to Hermitian operators acting on the Hilbert space, $\mathcal H$, containing the state vectors, $|\psi\rangle$ - the set of all eigenvectors of each operator can be used as a basis for $\mathcal H$. This includes (brushing a few things under the carpet) the position operator, $\hat x$.

Now imagine you are given a state, $|\psi\rangle$. If we change the basis for $\mathcal H$ to the one generated by the position eigenvectors, the components of $|\psi\rangle$ can be written $\langle x|\psi\rangle=\psi(x)$. So for a given value of $x$, $\psi(x)$ represents one particular component of the state vector $|\psi\rangle$. We essentially parameterise the components of $|\psi\rangle$ by $x$ and this produces a function of $x$.

There are some technical issues we are avoiding here, quantum mechanics is complicated. A more mathematical statement of what I've just written would be to say that $\mathcal H$ (which contains the $|\psi\rangle$'s) is isomorphic as a Hilbert space to the space of square integrable functions on an interval (which contains the $\psi(x)$'s): $$\mathcal H\cong L^2(\Bbb R). \tag{1}$$

Edit: Maybe this (crudely drawn) diagram helps explain roughly what we're doing here (it is not rigorous, only intended to help visualise it):

enter image description here

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  • $\begingroup$ in psi(x) what is the x? $\endgroup$ – STEMguy24 Nov 2 '20 at 18:17
  • $\begingroup$ Well, in wavefunction notation it denotes the $x-$coordinate. Imagine taking the state vector $|\psi\rangle$, which is just a column vector of (complex) numbers, and turning it on its side. Then you associate to each component of the vector, one point along the interval on which $\psi(x)$ is defined (which may be all of $\Bbb R$). So the very first component of the column vector becomes $\psi(0)$ and each consecutive component of $|\psi\rangle$ gets assigned the the next point along the number line until you reach $\psi(L)$. This is now a function defined on $[0,L]\subseteq R$, thats $\psi(x)$ $\endgroup$ – Charlie Nov 2 '20 at 18:28
  • $\begingroup$ yeah i dont understand $\endgroup$ – STEMguy24 Nov 2 '20 at 18:30
  • $\begingroup$ @STEMguy24 Does the picture I've added help at all? $\endgroup$ – Charlie Nov 2 '20 at 18:40
  • $\begingroup$ so is psi(X) probability you will find a particle in position x? What is H $\endgroup$ – STEMguy24 Nov 2 '20 at 18:42

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