I am in an undergraduate quantum mechanics course, and we are starting to use Dirac notation. I am a tad confused on exactly how all this works.

Having taken linear algebra, I am very comfortable with the idea of bases. So, I under stand that an operator transforms one vector to another:

$$ \hat Q |\alpha \rangle = |\beta \rangle$$

I am also comfortable with $$ | \alpha \rangle =\sum a_n| e_n \rangle \text{ where } a_n = \langle e_n | \alpha \rangle. $$ Also, the corresponding version for $\beta$:

$$ | \beta \rangle =\sum b_n| e_n \rangle \text{ where } b_n = \langle e_n | \beta \rangle $$

Then, the book defines

$$\langle e_m | \hat Q | e_n\rangle \equiv Q_{mn} $$

I understand this as well. The various $Q_{mn}$ values are simply the entries in the matrix that represents $\hat Q$.

Here is my issue...

My book claims that if a particle is in state $S(t)$,

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

where the vector $|x\rangle$ is "the eigenfunction of $ \hat x$ with eigenvalue $x$."

But the abstract form of this is a little perplexing to me. Can anyone sort of explicitly write out a matrix/vector representation of this? Also, is it even possible to write down $S(t)$? Because to my understanding it is a vector that has not been assigned a basis set. I have tried to look online and have also tried to play around with it myself, but haven't gotten very far.

  • $\begingroup$ Looks to me the position space wavefunction. Have a look at Sakurai. $\endgroup$
    – sbp
    Nov 2, 2017 at 0:04
  • $\begingroup$ You know, why don't you take x to range over 3 distinct points, 1,2,3. Weyl took N in his book, but 3 will suffice. So you have $|j\rangle$, i.e., $|1\rangle=(1,0,0)^T$, etc. See $\langle j|\hat{x}|i\rangle=j\delta_{ji}$. Take $\Psi (j)=\exp (-j^2)$. Write $|S\rangle= \sum_j \Psi (j) |j\rangle$. Etc... Now change bases by , e.g. a cyclic rotation of 1,2,3. It's not that bad... $\endgroup$ Nov 2, 2017 at 21:47

3 Answers 3


This is Dirac's formalism. It is a generalization to continuous basis, i.e., those enumerated by an index which takes values in a continuous set like $\mathbb{R}^n$.

In that setting you have that in the same way that $|e_n\rangle$ is a discrete basis which allows you to decompose

$$|\psi\rangle=\sum_{n}\langle e_n|\psi\rangle |e_n\rangle,$$

and with respect to which the closure relation holds

$$\sum_{n}|e_n\rangle\langle e_n|=\mathbf{1}$$

we suppose that we can have one basis $|x\rangle$ enumerated by some continuous parameter like $x\in \mathbb{R}$, such that we can decompose

$$|\psi\rangle=\int \langle x|\psi\rangle |x\rangle dx$$

with closure relation

$$\int|x\rangle \langle x|dx = \mathbf{1}.$$

The issue is that roughly if $A$ is one compact operator, the spectral theorem will ensure you have a discrete orthonormal basis of eigenvectors $|a_n\rangle$ such that $A|a_n\rangle = a_n |a_n\rangle$ (I assume non-degenerate for simplicity).

When $A$ is unbounded as often happens in QM, no such basis exists. But you suppose that the above sort of generalized basis do exist. So if $X$ is unbounded you assume that for every eigenvalue $x\in \sigma(X)$ the spectrum of $X$ there is a state ket $|x\rangle$ with $X|x\rangle = x|x\rangle$ and forming a basis.

Notice that whenever you have position and momentum operators $X,P$ you want to require $[X,P]=i\hbar$ and there is a theorem which ensures that at least one of them will be unbounded, so the thing above will be needed then.

This is important due to the postulates of QM. Observables are hermitian operators. The possible values to be measured are exactly the values on the spectrum, i.e., the "eigenvalues" and the states with definite value of the quantity are the eigenvectors, the value measured is then the corresponding eigenvalue.

It is then postulated that if $A$ is the observable with continuous basis $|a\rangle$ then $\rho(a) = |\langle a|\psi\rangle|^2$ is the probability density of finding the value of $A$ in the state $|\psi\rangle$ to be between $a$ and $a+da$.

You then connects this with wave mechanics. Consider a particle in one dimension. We have the observable $X$ corresponding to position. Let $|S(t)\rangle$ be the state at time $t$. As we know position can assume any possible value so $\sigma(X)=\mathbb{R}$. Let $x\in \mathbb{R}$, the corresponding generalized eigenvector is $|x\rangle$. The probability of finding the particle between $x$ and $x+dx$ is then $\rho(x) = |\langle x|\psi\rangle|^2$.

So making contact with wave mechanics, we see that $\Psi(x,t)=\langle x|S(t)\rangle$ indeed.

As a final remark, all the thing about generalized eigenvectors and continuous basis from Dirac's formalism is extremely useful and elegant, but it is non rigorous. In rigorous functional analysis there is no eigenvector for unboudned operators and these expansions are not defined. There is, though, one workaround that makes it all make sense, called the Gel'fand triplet approach.

  • $\begingroup$ Let's apply it to a simple example, say, the particle in a 1D well. For this the general solution is $\Psi(x,t) = A Sin(n k x)$. What, then, is $S(t)$? $\endgroup$
    – Ben
    Nov 2, 2017 at 0:32
  • 1
    $\begingroup$ Notice that you are giving one time independent state, but anyway, the thing is: when you have a probability amplitue $\psi(x)$ like this one, $\psi(x) = A\sin (nkx)$ you can immediately write down the ket $|\psi\rangle = \int \psi(x) |x\rangle dx$. It is one expansion in a basis. Think of $\psi(x)$ as the components. The ket is abstract and is good for abstract manipulations that sometimes makes it easier to find the state itself (see the harmonic oscilator example in Cohen's book), but actual computations are done specifying $\psi(x)$ and remembering $\psi(x) = \langle x|\psi\rangle$. $\endgroup$
    – Gold
    Nov 2, 2017 at 0:35
  • $\begingroup$ In |ψ⟩=∫ψ(x)|x⟩dx is |x⟩ basically the eigenfunctions of position? $\endgroup$
    – Ben
    Nov 2, 2017 at 0:52
  • $\begingroup$ Notice that boundedness is not sufficient for a discrete spectrum. The position operator in the infinite square well is a simple counterexample. The spectral theorem requires stronger hypotheses and as I understand it there's a healthy variety of approaches, with different resulting strengths of the theorem. $\endgroup$ Nov 2, 2017 at 6:58
  • 1
    $\begingroup$ That's a sufficient condition but it is not necessary, if I understand the theorem correctly. A simple example of a non-compact operator with an eigenvector basis is the harmonic oscillator. $\endgroup$ Nov 2, 2017 at 14:06

A simple (but rather non-rigorous) way to think about this:

Your Hilbert space $\mathcal{H}$ is formed by square-integrable functions on the real line, that is elements of $\mathcal{H}$ are given by maps $\psi: \mathbb{R} \rightarrow \mathbb{C}$ such that $\int_{-\infty}^{\infty} dx \,|\psi(x)|^2$ is finite, for instance a Gaussian, $\psi(x) = e^{-x^2/2}$.

Each "ket" you write down $|\psi\rangle$ is representing some map $\psi: \mathbb{R} \rightarrow \mathbb{C}$.

Inner products of two such kets $|\psi\rangle, |\phi \rangle$ is just given by $\langle \psi| \phi\rangle = \int_{-\infty}^{\infty} dx\, \psi^*(x) \phi(x)$ where $\psi(x), \phi(x)$ are the maps corresponding to those kets.

A simple way to think about the position eigenket $|f_y\rangle$ is just that it represents the "function" $f_{y}(x) = \delta(x-y).$ These are precisely the "eigenfunctions" of the position operator $\hat{x} : \mathcal{H} \rightarrow \mathcal{H}$ which takes $\psi(x) \rightarrow x\ \psi(x)$. That is $f_{y}(x)$ are the only functions which solve the eigenvalue equation $\hat{x} f_y(x) = \lambda f_y(x)$ with eigenvalue $\lambda = y$. The reason for the quotes around the word function and eigenfunction is that $f_y(x)$ are not square-integrable, that is, they aren't really in $\mathcal{H}$.

Making these notions precise and rigorous is the subject of functional analysis.

  • $\begingroup$ So I am gathering that the following are equivalent: $\Psi(x,t) = \langle x | \psi \rangle = \int \delta (x-y) \psi dy $ Is this correct? $\endgroup$
    – Ben
    Nov 2, 2017 at 2:32
  • $\begingroup$ Drop the $t$ and yeah, it becomes a trivial identity. $\endgroup$ Nov 2, 2017 at 3:18
  • $\begingroup$ Oops. I meant: Ψ(x,t)=⟨x|ψ⟩=∫δ(x−y)ψ(t)dy $\endgroup$
    – Ben
    Nov 2, 2017 at 3:24
  • $\begingroup$ But in the evaluation of that integral... $\int \delta(x-y) \psi(t) dy = \psi(t) \int \delta(x-y)dy = \psi(t)$ Which means... $\Psi(x,t) = \psi(t)$. But how can this be? $\endgroup$
    – Ben
    Nov 2, 2017 at 3:26
  • 1
    $\begingroup$ No, these are all stationary states. It should read $\langle f_y | \psi \rangle = \int dx \delta(x-y) \psi(x) = \psi(y)$. If you insist on time evolving the state, you should write the ket $|\psi(t) \rangle$ which then represents the time-evolved wave function $\psi(x,t)$. You would then get $\langle f_y | \psi(t) \rangle = \int dx \delta(x-y) \psi(x,t) = \psi(y,t)$ $\endgroup$ Nov 2, 2017 at 3:28

The important thing to have in mind (and which is not always stressed in books) is that the wave function $\psi$ is actually the collection of the coefficients representing the state of the system in a given basis.

The vector $|S\rangle$ is a completely abstract element of a vector space which by means of a postulate shall represent completely state of the system. Let us say you have an observable $A$ with eigenvalues $a$ and corresponding eigenstates $|a\rangle$, $$A|a\rangle=a|a\rangle,$$ then the state $|S\rangle$ can be spanned in the basis $\{|a\rangle\}$, $$|S\rangle=\sum_a\psi(a)|a\rangle.$$ The collection of coefficients $\psi(a)$ is the wave function associated to the state $S$ in the basis defined by the observable $A$. If the observable $A$ have continuous eigenvalues, then the sum above is replaced by an integral and the collection $\{\psi(a)\}$ has infinite elements and is actually a continuous function on the variable $a$.

Now let us consider the concrete example of the state of the system in the basis defined by position. If the operator $\hat x$ has eigenstate $|x\rangle$ with eigenvalue $x'$, then $$\hat x|x\rangle=x'|x\rangle,$$ Since the wave function of this state is just $\psi(x)$, the above equation can be written as $$x\psi(x)=x'\psi(x)\Rightarrow (x-x')\psi(x)=0.$$ The last equation says that $\psi(x)$ is non-zero only when $x=x'$, i.e., $$\psi(x-x')=\delta(x-x').$$ Now consider the inner product of an arbitrary state $$|S\rangle=\int\psi(x)|x\rangle dx,$$ with a position eigenstate $|x'\rangle$, $$\langle x'|S\rangle=\int\delta(x-x')\psi(x)dx=\psi(x').$$ Since $x'$ is arbitrary we can say that $$\langle x|S\rangle=\psi(x),$$ which is the wave function in the position representation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.