I am not sure how the wave function can be viewed as quantum state.

I begin with the eigen-equation

$$A|\psi\rangle = a|\psi\rangle$$

If $A$ is a $n$ dimensional matrix with different eigenvalues $a$, then there will be $n$ eigenvectors.

If I do orthonormalization to these eigenvectors, then there will be $n$ basis vectors.

If the $|\psi\rangle$ is vector in quantum state space, then this is the $n$ dimensional quantum state.

Next, more specific consideration. The Schrodinger equation, $$H|\psi\rangle=E_n|\psi\rangle$$ with at the $n$ energy eigenvalue and $H$ is Hamiltonian (not in form of the matrix like above). Normally it is solved in the position state, that will be $$H\Psi_n(x) = E_n\Psi_n(x)$$ where $\Psi_n(x) = \langle x|\psi\rangle$ is wave function of an $n$ energy eigenvalue.

In this form, the wave function is just a continuous coefficient of $|x\rangle$.

How does this $\Psi_n(x)$ become quantum state ?

Can I say that because there are $n$ wavefunctions on the same position space $|x\rangle$ then they form the Hilbert space and can be viewed as quantum state?

$$\int {\Psi_m}^*(x) \Psi_n(x) dx =<\Psi_m(x)|\Psi_n(x)>= \delta_{m,n}$$


2 Answers 2


A wave function can be viewed as describing a quantum state in the same way that the components of a vector can be viewed as describing the vector itself. Yes, $\psi(x_0)=\langle x_0\vert\psi\rangle$ is the numerical value of $\psi(x)$ at $x=x_0$ but this numerical value is basically the component of the vector $\vert\psi\rangle$ along the basis vector $\vert x_0\rangle$ (see this question), just like $\langle \hat k\vert \vec r\rangle=r_k$.

If you prefer to work in a momentum basis, then $\psi(p)=\langle p\vert\psi\rangle$ and the change of basis formula is contained in the overlap $$ \langle p\vert x\rangle = \frac{e^{-i px/\hbar}}{\sqrt{2\pi \hbar}} $$ so that $$ \langle p\vert\psi\rangle = \int dx \langle p\vert x\rangle \langle x\vert\psi\rangle $$ just like a change of basis is given by $$ r_a =\langle \hat a\vert\vec r\rangle =\sum_k \langle \hat a\vert \hat k\rangle \langle \hat k\vert \vec r\rangle=\sum_k A_{ak}r_k\, . $$

Asher Peres in

Peres, Asher. "What is a state vector?." American Journal of Physics 52.7 (1984): 644-650.

quotes Ekstein as stating that "states are images of procedures by which a system is made to interact with a macroscopic apparatus". The reference to the original Ekstein paper is Ekstein, H. "Presymmetry." Physical Review 153.5 (1967): 1397.


The wavefunctions themselves are simply the coefficients in the expansion of the quantum mechanical state vector in the position basis, as you already said:

$$\Psi(x, \ t) \ = \ \langle x | \Psi \rangle$$

I'm not sure if a Hilbert space is the right word to use, as the wavefunctions themselves are not vectors, however, any other wavefunction in our possible state space can be expressed as a linear combination of the stationary state wavefunctions of the Hamiltonian.

I'm not totally sure what you mean when you ask how the wavefunction becomes the quantum state. The point of the set of stationary state wavefunctions in the position basis is that we can write the quantum mechanical state vector as a linear combination of them.

For instance, consider the position basis $|x\rangle$. We can write some state vector as a linear combination of position eigenstates. Remember that:

$$\displaystyle\sum_{i} \ |v_i\rangle \langle v_i| \ = \ \mathbb{I}$$

For any orthonormal basis as:

$$\Big( \displaystyle\sum_{i} \ |v_i\rangle \langle v_i| \Big) |v\rangle \ = \ \displaystyle\sum_{i} v_i |v_i\rangle \ = \ |v\rangle$$

Now, this also holds for a continuous sum (an integral), so we have:

$$|\Psi\rangle \ = \ \mathbb{I} |\Psi\rangle \ = \ \displaystyle\sum |x\rangle \langle x | \Psi \rangle \text{d}x \ = \ \displaystyle\sum \Psi(x, \ t) |x\rangle \text{d}x$$

Cool, so we can expand the state vector in terms of a basis of eigenstates. But we can also expand the wavefunction in terms of another basis! For example, we can then expand the position wavefuntion in terms of the energy eigenstate basis. Thus, we have:

$$\langle x | \Psi \rangle \ = \ \langle x | \mathbb{I} | \Psi \rangle \ = \ \langle x | \Big( \displaystyle\sum_{n} \ |E_n\rangle \langle E_n | \Big) | \Psi \rangle \ \ = \ \displaystyle\sum_{n} \ \langle x |E_n\rangle \langle E_n | \Psi \rangle \ = \ \displaystyle\sum_{n} c_n \Psi_n(x, \ t)$$

So we can write the state vector as the wavefunction, and we can write the wavefunction as a linear combination of stationary states, with each of the $c_n$ coefficients corresponding to the probability amplitude of the specific energy eigenstates, thus telling us "how much" of each $\Psi_n(x, \ t)$ is in $\langle x | \Psi \rangle$.

So to answer your question, the wavefunctions are not quantum states, they are components of the quantum states that tells us certain information about the quantum state represented in a particular basis, but we can still expand/express each one as linear combinations of vector/functions in another basis.

  • $\begingroup$ Suppose $H|\psi> = E|\psi>$ with H is $n$ dimesion matrix. Solve it like normal eigen-equation, then there will be the $n$ dimesion of basis of energy correspond to eigen value $E_n$. The eigen vector $|\psi>$ will be $|\psi> = \Sigma_i^n c_i |E_i>$. Next I express each of them in term of position basis, they will be $|\psi> = \Sigma_i^n c_i \int <x|E_i>|x>dx = \Sigma_i^n c_i \Psi_i(x)|x>$. right? $\endgroup$ Commented Sep 25, 2019 at 5:19
  • $\begingroup$ Not quite, the energy eigenstates of $H|\psi> = E|\psi>$ are the set $|E_n>. You seem to try to express the energy eigenvalues in terms of themselves in the first equation. $\endgroup$ Commented Sep 25, 2019 at 11:12
  • $\begingroup$ the correct way to do this would be to find the energy eigenvalues $H|E_n\rangle \ = \ E_n|E_n\rangle$. If we want to find the stationary position wavefunctions, we expand these eigenvectors in terms of the position basis: $|E_n\rangle \ = \ \displaystyle\int \ \langle x | E_n \rangle |x\rangle dx \ = \ \displaystyle\int \ \psi_n(x) \ |x\rangle dx$. $\endgroup$ Commented Sep 25, 2019 at 12:10
  • $\begingroup$ In order to recover the full state vector from the wavefunction, we need to know the initial wavefunction for the system. To find the time-dependent wavefunction, we have to express the initial wavefunction as a linear combination of the energy stationary position wave functions, then add time-dependence, then integrate over this wavefunction with respect to all the position eigenstates. $\endgroup$ Commented Sep 25, 2019 at 12:15

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