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So I am reading through Griffiths and I am a little stuck. He says the state of a quantum system can be represented by some state $\left|\mathscr{s}(t)\right>\in \mathcal{H}$. However we can express this state in any number of basis. All of this makes sense. Where he loses me though is he then goes on to say that our wave function $\Psi(x,t)$ is actually the coefficient of the expansion of $\left|\mathscr{s}(t)\right>$ in the basis of position eigenfunctions.

Namely: $$\Psi(x,t)=\left<x|\mathscr{s}(t)\right>$$

Hasn't $\Psi(x,t)$ been an element of our Hilbert Space $L^2[a,b]$ this whole time? so why is our wave function now a scalar instead of a vector (element of $\mathcal{H}$)?

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    $\begingroup$ A function is an infinite dimensional vector. $\endgroup$ – Gradient137 Nov 19 '18 at 0:26
  • $\begingroup$ But the inner product of two vectors i.e. $\left<x|\mathscr{s}(t)\right> \in \mathbb{C}$ so by the definition that Griffiths gives wouldnt $\Psi(x,t)$ be a scalar? $\endgroup$ – Alex Sampson Nov 19 '18 at 0:34
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    $\begingroup$ The function $\Psi$ evaluated at the point $x$ is a complex number, but this is the same thing as saying "the $z$-component of a position vector is a real number". Since the $|x\rangle$ vectors are the basis vectors of the Hilbert space, what you are doing in that inner product is simply selecting a particular component of a vector. The function itself (i.e. the collection of all of its "components") is still a vector. $\endgroup$ – probably_someone Nov 19 '18 at 0:40
  • $\begingroup$ This won't answer your question, but said in another way: consider $u,v \in \mathbb{R}$. $2+3 = 5$ but if I write $u + v$ then I have an expression, whose evaluate depends on what $u$ and $v$ are. Therefore, let me define a function $f: (u,v) \mapsto (u+v)$ with $f(u,v) := u + v$. Likewise, me naively looking at your first equation from a math perspective, we have the same thing. $\left<1 \text{meter}|s(2 \text{seconds})\right> \in \mathbb{C}$. And now here's where I'm going to guess what is happening because I don't know the physics. $x$ means something and $t$ means something, but both $\endgroup$ – DWade64 Nov 19 '18 at 14:30
  • $\begingroup$ are variables so we define a function $\Psi : (x,t) \mapsto \left<x|s(t)\right>$ or $\Psi(x,t) := \left < x|s(t)\right >$. I remember reading this textbook before. I'm pretty sure the symbol $\Psi$ is being overloaded [often done in physics. Overloading is like $x = x(t)$ where $x$ is meant to stand for the name of a function (defined somewhere else on your sheet of paper) and also a variable $x$. It's better to write something like $y = f(x)$ where $f$ is a function (defined somewhere) and $y = f(x)$ is an equation satisfied by some $(x,y)$ pair] Because Griffiths recognized this, he $\endgroup$ – DWade64 Nov 19 '18 at 14:33
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It's still a vector, just with an uncountably infinite number of components (one component for each value of $x$).

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  • $\begingroup$ But isnt the inner product of two vectors a scalar? So with Griffiths definition since: $$\Psi(x,t)=\left<x|\mathscr{s}(t)\right>$$ Then $\Psi(x,t)$ is the inner product of two vectors in $\mathcal{H}$ then wouldnt that mean that $\Psi(x,t)$ is a scalar? I figured that it must still be a vector, but I dont see how it is from the definition in Griffiths. $\endgroup$ – Alex Sampson Nov 19 '18 at 0:40
  • $\begingroup$ @AlexSampson What you're missing here is that though the output of the function $\Psi$ is a scalar, that does not mean that the function itself is a scalar. This is analogous to many other situations in linear algebra; for example, a 2x2 matrix is a map that inputs a 2D vector and outputs a 2D vector, but is not itself a 2D vector. $\endgroup$ – probably_someone Nov 19 '18 at 0:43
  • $\begingroup$ @AlexSampson For another example, the components of the velocity vector are real numbers. That does not mean that the velocity vector itself is a real number. The inner products you're looking at are the components of a vector. $\endgroup$ – probably_someone Nov 19 '18 at 0:45
  • $\begingroup$ oh so in his definition is the $\Psi(x,t)$ is the $i^{th}$ component of the vector $$\left|\Psi(x,t)\right>=\sum_i \left|x_i\right>\left<x_i|\mathscr{s}(t)\right>$$ where $\left|x_i\right>$ are all the different eigenfunctions of the position operator. Is that the correct reasoning. (hopefully that sum means what i want it to mean i'm just getting used to Dirac notation.) $\endgroup$ – Alex Sampson Nov 19 '18 at 0:53
  • $\begingroup$ @AlexSampson Yes, that's the idea. In the interest of precision, $i$ runs over an uncountable set in your notation. $\endgroup$ – probably_someone Nov 19 '18 at 0:55

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