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My understanding is that $\psi(\vec{r}, t)$ and $|\psi(\vec r,t)\rangle$ are the same thing yet one expressed as a wave function and the other expressed as a vector in the Hilbert space. Is this true? Or is there a deeper difference between the two notations?

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    $\begingroup$ $\psi(\vec{r},t)=\langle \vec{r}|\psi(t)\rangle$. $\endgroup$
    – velut luna
    Commented Apr 21, 2017 at 11:49
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    $\begingroup$ $|\psi\rangle = \int \psi(\vec{x},t)|\vec{x}\rangle d^3\vec{x}$ $\endgroup$
    – gautampk
    Commented Apr 21, 2017 at 11:53
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    $\begingroup$ Which page does Griffiths write $|\psi(\vec r,t)\rangle$? $\endgroup$
    – Qmechanic
    Commented Apr 21, 2017 at 11:59
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    $\begingroup$ related: physics.stackexchange.com/q/65794 $\endgroup$
    – Kyle Kanos
    Commented Apr 21, 2017 at 12:15
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    $\begingroup$ Yes, it is a wrong notation. It doesn't represent any mathematical object. It combines the notation of two very different ways of representing a quantum state. $\endgroup$
    – garyp
    Commented Apr 21, 2017 at 13:04

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$\psi (\vec{r},t) $ is like you said, just a way to express the vector $|\psi (t)\rangle $ in 'position space', mathematically expressed like is written in the comments:

$$ \psi (\vec{r},t) = \langle \vec{r} | \psi(t) \rangle = \int \delta(r'-r)\psi(t) d^3 r $$

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Velut Luna gives the main answer. One can see this because we have the probability expectation $1~=~\langle\psi(t)|\psi(t)\rangle$ and with the completion sum $\mathbb I~=~\int d^3r|\vec r\rangle\langle \vec r|$ we then have $$ 1~=~\langle\psi(t)|\psi(t)\rangle~=~\langle\psi(t)|\left(\int d^3r|\vec r\rangle\langle\vec r|\right)|\psi(t)\rangle~=~\int d^3r\langle\psi(t)| \vec r\rangle\langle\vec r|\psi(t)\rangle. $$ In the wave function form we have unity of probability as $$ \int d^3r\psi^*(\vec r,t)\psi(\vec r,t). $$ the identification is obvious.

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It is convenient to think of $\vert\psi\rangle$ as a vector with components $\langle x\vert\psi\rangle=\psi(x)$ for various values of $x$. If you imagine discrete rather than continuous values of $x$, then the vector $\vert\psi\rangle$ would be the infinite column vector $$ \left(\begin{array}{c} \vdots \\ \psi(x_{n-2})\\ \psi(x_{n-1})\\ \psi(x_{n})\\ \psi(x_{n+1})\\ \psi(x_{n+2})\\ \vdots \end{array}\right) = \left(\begin{array}{c} \vdots \\ \langle x_{n-2}\vert \psi\rangle \\ \langle x_{n-1}\vert \psi\rangle \\ \langle x_{n}\vert \psi\rangle \\ \langle x_{n+1}\vert \psi\rangle \\ \langle x_{n+2}\vert \psi\rangle \\ \vdots \end{array}\right) $$ obtained by decomposing the vector $\vert\psi\rangle$ on the basis of states $\{\ldots, \vert x_{n-2}\rangle,\vert x_{n-1}\rangle,\vert x_{n}\rangle,\vert x_{n+1}\rangle,\vert x_{n+2}\rangle\ldots\}$

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There is a distinction that is, in my opinion, quite deep and subtle about the two different notations. The second is far more versatile than the first, and universally usable in quantum mechanics (while the first is not). To clarify, the two notations are:

  • A complex-valued function $\psi(\cdot)$, with some space - e.g. $\mathbb{R}^d$ or $\{\uparrow,\downarrow\} \times \mathbb{R}^d$ - as domain.
  • A vector in a Hilbert space $\psi\in\mathscr{H}$ (or $\lvert \psi\rangle$ if you prefer, I will use the former).

The explanation why the second is more universal goes as follows. We know that any "reasonable" physical quantum system can be described mathematically by a so-called non-abelian C*-algebra, that represents the set of (noncommutative) observables of the system. In turn, any C*-algebra can be represented by a set of linear transformations on some Hilbert space.

Now, the C* algebra of $N$ non-relativistic quantum particles in $d$ space dimensions has a unique irreducible representation up to unitary equivalence (without entering into details, irreducible repr. are the relevant ones). Such representation is the algebra of (bounded) operators on the space $L^2(\mathbb{R}^{dN})$, and the self-adjoint (unbounded) operators $(x_1,\dotsc,x_N)$ and $(-i\nabla_1,\dotsc,-i\nabla_N)$ represent the position and momentum operators of each particle. Therefore, it is in this case natural to write an element of the Hilbert space of non-relativstic quantum mechanics as a (wave)function $\psi(x_1,\dotsc,x_N)\in L^2(\mathbb{R}^{dN})$.

The latter, however, ceases to be true for relativistic quantum mechanics: there are infinitely many irreducible inequivalent representations for the algebras of quantum field observables, and in particular there is no unique natural description in terms of a wavefunction(al). In this case, therefore, the functional notation $\psi(\cdot)$, even in representations where it is viable, would be rather ambiguous and not as "universal" as it is in non-relativstic quantum mechanics.

In conclusion, as long as non-relativstic quantum mechanics is concerned, the distinction is almost only aesthetic, while for more general (relativistic) theories one should essentially abandon the idea of a "wavefunction" altogether and consider more abstract representations of the canonical commutation relations, for which the notation $\psi(\cdot)$ could make no sense (while $\psi$ still make sense).

Finally, let me also anticipate some possible comments. It is true that all the separable infinite-dimensional Hilbert spaces are isomorphic, but there are still inequivalent representations of the C* algebra in relativistic systems. Given the vacuum vector $\Omega$ in a given separable representation $\mathscr{K}$, it is indeed possible to map it to a wavefunction in e.g. $L^2(\mathbb{R})$, but it is impossible to tell what the field operator would be in the latter (and so the map is useless).

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