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.

  • $\begingroup$ Concerning just the title: I think $\psi$ and $\left |\psi \right >$ is confusing notation to a mathematician. Let $\left | \psi \right >$ be a vector in Hilbert space (where vector just means an element of a vector space, here Hilbert space). We already used up the symbol $\psi$ in $\left | \psi \right >$. Therefore I shouldn't use it again (which would overload it with multiple meanings). Let $f$ be a function defined as $f: (x,t) \mapsto \left < x| \psi(t)\right >$. So the output of $f$ gives the component of $\left | \psi \right >$ along a particular $\left | x\right>$. However notice how $\endgroup$
    – DWade64
    Nov 22, 2018 at 14:45
  • 1
    $\begingroup$ physics people give functions their names. If $g$ is a function of time and outputs a position $x$, then $g(t)$ is often written $x(t)$. Physics people read this "position as a function of time" so they give the function name $g$ as $x$ so they can read out loud "position as a function of time." But this is overloading because usually you'd have $x = g(t)$, but with this overloading we have $x = x(t)$. So back to my first comment. Instead of using $f$, physics people overload the symbol $\psi$ so they have $\psi(x,t)$ and they can read "component of $\left |\psi \right >$ as a function of x,t" $\endgroup$
    – DWade64
    Nov 22, 2018 at 14:48
  • $\begingroup$ LPT: Use \rangle to get more appropriate-looking ket's: $\vert\psi\rangle$ gives $\vert\psi\rangle$. $\endgroup$
    – Kyle Kanos
    Nov 23, 2018 at 12:18
  • $\begingroup$ Re: your edit today, consider when you use ordinary finite vectors versus triplets of their components. Often the problem chooses the language, and, I have to tell you, the abstract ket notation has been getting increasingly attractive by the decade for many of us... Dirac really cleaned things up in his book. $\endgroup$ Feb 7, 2019 at 1:52

2 Answers 2


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


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.

  • $\begingroup$ So does the ket $|\psi \rangle$ live in an equivalence class of Hilbert spaces, while the functions $\psi(x)$ and $\psi(k)$ live in what mathematicians call a "Hilbert space"? $\endgroup$ Nov 25, 2018 at 15:01
  • $\begingroup$ @EricDavidKramer The vector $\left|\psi\right>$ lives in Hilbert Space and also $\psi\left(x\right)$ and $\psi\left(k\right)$. The latters are just different ways to define sets (e.g. set of all functions such that...) which have the same Hilbert Space structure. Those are just representations. $\endgroup$
    – eranreches
    Nov 25, 2018 at 23:06

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.

  • $\begingroup$ I don't have the math to fully understand this but consider a state vector $\left |\psi(t)\right>$ and the function $\psi$ showing it's projection $\psi:(x,t) \mapsto \left<x|\psi(t)\right>$ (which I could have written as $\psi(x,t) := \left<x|\psi(t)\right>$). While $\psi(x,t)$ lives in the space of square-integrable functions (which is a Hilbert space). You're saying that $\left | \psi \right >$ also lives in a Hilbert space but not the same Hilbert space as $\psi(x,t)$ right? However, either or, can be used to talk about a system (like a function and it's inverse function are "equivalent")? $\endgroup$
    – DWade64
    Nov 22, 2018 at 15:49
  • $\begingroup$ @DWade64 I am saying that, as long as only hilbert space structure plays a role, all ways of identifying the space are equivalent. I mean that you can see any element as an abstract vector $|\psi\rangle$, as a square integrable function $\psi(x)$, as a square summable sequence $(\psi_n)_{n\in \mathbb{N}}$, etc. This is because there is always a map (structure preserving and bijective) identifying the spaces (and thus all their objects accordingly). What is important is to consider the observables, i.e. the physical content of the theory, and then not all representations are equivalent. $\endgroup$
    – yuggib
    Nov 22, 2018 at 16:10
  • $\begingroup$ I don't have a large math background. The way I was thinking about my first comment was: consider a real vector space of position vectors. Let it also have an inner product. Then $(1,1,1) \cdot (2,3,4) = 2 + 3 + 4 = 9$. But 9 is just a number. While (1,1,1) and (2,3,4) are elements of the vector space, 9 wouldn't be considered an element of the vector space (maybe 9 would be considered an element of the field of the vector space?). I was using this to make the analogy $\left | \psi \right >$ are like the vectors (1,1,1) and (2,3,4) while $\psi(x,t)$ is like the number 9 but I don't know $\endgroup$
    – DWade64
    Nov 22, 2018 at 16:52
  • $\begingroup$ @DWade64 A function is not a simple number. You can think of a function $f\left(x\right)$ as an infinite vector like $\left(...,f\left(-2{\rm d}x\right),f\left(-{\rm d}x\right),f\left(0\right),f\left({\rm d}x\right),f\left(2{\rm d}x\right),...\right)$. $\endgroup$
    – eranreches
    Nov 22, 2018 at 19:13
  • $\begingroup$ @eranreches Yeah, I was likening $9$ to something like $\psi(1,2)$ in my head which I should have been more explicit about $\endgroup$
    – DWade64
    Nov 22, 2018 at 20:12

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