Link between wave functions $\psi$ and state vectors $|\psi \rangle$ in quantum mechanics? [duplicate]

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• In quantum mechanics, is $|\psi\rangle$ equal to $\psi(x)$? 6 answers

I'm currently following a course in quantum mechanics that uses Griffith's textbook.

Griffiths shows that wave functions are members of a Hilbert space. Since this is an abstract vector space, we can assign a state vector $|{\psi}>$ to each wave function $\psi(x)$. Since our course on linear algebra runs pretty deep, this I understand.

Griffiths subsequently defines the inner product between two state vectors $<\phi|\psi> = \int\phi^*(x)\psi(x)dx$, and he derives all sorts of fun things from this inner product; that the function $\psi(x)$ corresponding to $|\psi>$ is the wave function as seen previously in wave mechanics, and that its Fourier transform determines the probability density of momentum.

At this point, something happens in both Griffiths and my course that I don't quite comprehend. Both my instructor and the book mention that at this point, the vectors will take precedence above the functions; all quantum mechanical information is from now on fully and completely recorded into the vector. The functions are simply a representation of this vector; one of the many possible representations. Am I correct in saying this?

But Griffith's proof that the coefficients in the position basis of $|\psi>$ form the values of the function $\psi(x)$ is completely based on the definition $<\phi|\psi> = \int\phi^*(x)\psi(x)dx$. How should I then read this? Does he assume that there is a function representation $\psi(x)$ corresponding to the state vector $|\psi>$ (which is in his right, since $L_2$ is isomorphic to the vector space), to subsequently derive the properties of this function?

To me it just seems really strange that we have these abstract state vectors, only to define the inner product in terms of one very specific representation of that vector.

Edit 6 February 2019: I can (for obvious reasons) not unmark this question as a duplicate, but I can "edit to explain why my question hasn't been answered before." My question is not about whether the ket psi and the wave function are the exact same--I have written in my question that I understand that the wave function is a representation of the ket (in other words, I've used the answer to the duplicate question to ask my own question). My question is why in so many derivations, it appears that the wave function formulation takes precedence (in e.g. the definition of the inner product). In the answers below, it has been explained that there is no precedence, and that this is simply a shortcut in the theory.

Answer 1

The functions are simply a representation of this vector; one of the many possible representations. Am I correct in saying this?

Yes. You are probably familiar with the position basis $\left|x\right>$ which satisfies

$$\hat{x}\left|x\right>=x\left|x\right>$$

The wavefunction is then merely a representation of the state vector in the position basis

$$\psi\left(x\right)=\left<x| \psi\right>$$

Try to use the completeness relation given below and the fact that $\left<x|k\right>=\frac{1}{\sqrt{2\pi}}e^{ikx}$ to show that the momentum representation of the state vector is the Fourier transform of the wavefunction.

But Griffith's proof that the coefficients in the position basis of $\left|\psi\right>$ form the values of the function $\psi(x)$ is completely based on the definition $\left<\phi|\psi\right> = \int\phi^{\ast}\left(x\right)\psi\left(x\right){\rm d}x$. How should I then read this? Does he assume that there is a function representation $\psi(x)$ corresponding to the state vector $\left|\psi\right>$ (which is in his right, since $L^{2}$ is isomorphic to the vector space), to subsequently derive the properties of this function?

You don't have to define the inner product like this. The inner product is an invariant, and it does not depend on the specific choice of basis. The specific formula from Griffith's is just a consequence of the basis $\{\left|x\right>\}$ being complete. To elaborate, just use the completeness relation

$$\hat{I}=\int{\rm d}x\left|x\right>\left<x\right|$$

to get

$$\left<\phi|\psi\right>=\left<\phi\right|\hat{I}\left|\psi\right>=\left<\phi\right|\int{\rm d}x\left|x\right>\left<x|\psi\right>=\\=\int{\rm d}x\left<\phi|x\right>\left<x|\psi\right>=\int{\rm d}x\phi^{\ast}\left(x\right)\psi\left(x\right)$$

Nevertheless, there is something beneficial with the position basis: in it the Hamiltonian operator assumes a simple form which is the usual Schrödinger equation in term of wavefunctions.

Answer 2

The point is that, if you forget completely about the physics, and focus on the mathematical structure of vector spaces that are complete w.r.t. the topology given by an inner product, then they are all the same, once the "dimension" is fixed. All finite dimensional Hilbert spaces of dimension $d$ are isomorphic to $\mathbb{C}^d$ with the standard inner product, and all separable infinite dimensional Hilbert spaces are isomorphic to $\ell^2(\mathbb{N})$ (square summable sequences), or equivalently $L^2(\mathbb{R}^{n})$ (square integrable functions), with the standard inner products.

Therefore, at this preliminary abstract and mathematical level, all the information is indeed encoded in the "vector space" structure, and in the fact that it has an inner product w.r.t. which it is complete. If you want to prove things that only make use of the linear structure, of the inner product, and maybe of completeness and separability, then you can do it in any Hilbert space you like (e.g. $\ell^2$) and then translate the result to all other Hilbert spaces using the isomorphism.

However, there should be some important physical difference between the separable Hilbert space describing a single particle of spin zero in a one-dimensional box, and the separable Hilbert space describing a fermionic Dirac field acting on $3+1$ dimensional spacetime, even if they are abstractly "isomorphic".

The difference is given by physical observables. The important observation is that the aforementioned isomorphism between Hilbert spaces does not make physical sense, because it does not tell anything about the quantum observables, represented as operators acting on such Hilbert spaces. In fact Hilbert spaces are important for the quantum description of a physical system as long as it is clear which representation they carry of the set of quantum observables (such collection is in fact a C*-algebra that can always be represented as an algebra of bounded operators on some Hilbert space).

Two Hilbert spaces are "equivalent" in describing a given quantum mechanical system only if there is a (unitary) isomorphism between the respective representations of quantum observables. If else, they are not equivalent from the physical standpoint. In quantum mechanics (describing one or a finite number of $d$-dimensional particles), equivalent representation are given, for example, by the position and momentum representations, but also by the Fock or Bargmann-Fock representations (in fact all irreducible representations of observables are equivalent). In quantum field theories (describing a possibly infinite number of relativistic particles), there are many important representations that are not equivalent, even if they describe the same type of theory (e.g. a scalar field of mass $m$). A typical example are the representations corresponding to the free and interacting vacuum of a scalar field theory.