I am trying to do problem 2.4 in the book "Quantum field theory for the gifted amateur". I have a math background but little training in physics.

I am asked to use the identity $$\langle x \mid p \mid \psi \rangle = - i h \frac{d}{dx} \langle x \mid \psi \rangle.$$

My questions:

  1. what is the meaning of the $x$ appearing in the "bra". I understand that $\psi$ is a wave-function and $\hat{x}$ is the position operator. Is $x$ supposed to be a function? But this seems weird since it's obviously not in $L^2$ so the inner product doesn't seem to make sense.

  2. why is this identity true. (Please feel free to suggest a reference).


2 Answers 2


The bra (one form) $\langle x |$ or its corresponding ket vector $|x\rangle$ are "eigenstates" of the position operator $\hat{x}$ with eigenvalue $x$. I use the quotes because, in position or momentum co-ordinates, they are distributions, and not members of $L^2$. The momentum and position observables have no eigenstates in $L^2$, as discussed in my answer here. In position $x^\prime$ co-ordinates, $|x\rangle$ is the Dirac delta $\delta(x^\prime-x)$. Naturally, this same behavior holds for any operator with a continuous spectrum in the co-ordinates that transform the operator to a simple multiplication operator i.e. $\hat{x}(\psi(x)) = x\,\psi(x)$.

For your identity, see Emilio Pisanty's short explanation here.

  • $\begingroup$ Just to clarify: in your first sentence, the eigenvector x refers to "the eigenvector having eigenvalue x"? $\endgroup$
    – user142700
    Oct 8, 2017 at 0:41
  • $\begingroup$ @user142700 Yes. I distinctly recall writing "eigenvalue", but it seems my fingers were being a bit rebellious this morning. Fixed now $\endgroup$ Oct 8, 2017 at 1:04

The question is delicate from a mathematicaly rigorous point of view. What Physicists do is the following:

  1. States are described by elements of a Hilbert space called state space $\mathcal{E}$. One denotes the elements of $\mathcal{E}$ as $|\varphi\rangle$. On the space $\mathcal{E}$ there acts hermitian operators which represent physical quantities associated to the system, they are called observables. Given the particular system, we assume $\mathcal{E}$ has the necessary observables acting on it.

  2. The spectrum of one observable comprises the set of all possible valued to be measured. Now here comes the things: one assumes that every observable has a basis of eigenstates. When the observable is bounded the basis really exists as a discrete orthonormal basis $|\varphi_i\rangle$, which allows the decomposition of any state as $$|\varphi\rangle=\sum_{i=1}^\infty a_i |\varphi_i\rangle.$$ On the other hand, even when the obserable is unbounded, physicists assume a generalized continuous basis exists, i.e., that there is a continuous set of states $|x\rangle$ such that one can decompose $$|\varphi\rangle=\int\varphi(x)|x\rangle dx.$$ One also assumes here the completeness relation holds in this generalized sense $$\int|x\rangle\langle x| dx = \mathbf{1}.$$

  3. In the case of a particle which has position, for example, one assumes there is $\mathcal{E}$ with the position operator $X$ and the corresponding position basis $|x\rangle$ such that $X|x\rangle = x|x\rangle$ and such that the above decompositions can be done. Now the momentum operator $P$ should act as the generator of spatial translations and hence it should satisfy the equation you post.

It turns out that it is known that the generalized eigenvectors do not exist. The generalized eigenvectors exists only when you go into the so-called Gel'fand tripled (or Rigged Hilbert Space) formalism.

Nonetheless physicists assumes this can be done. So for your two points: (1) the basis $|x\rangle$ exists by assumption, while $\langle x|$ is the dual of $|x\rangle$, in the sense that $\langle x | (|\varphi\rangle) = (|x\rangle,\varphi\rangle)$. The wave function is the projection of the state on the position basis, so that $\varphi(x)=\langle x|\varphi\rangle$. (2) The identity is true, because one assumes there is a momentum operator satisfying it.

This is the traditional approach. But let's elaborate on some things. First of all, notice from (1) I wrote, that one assumes there is a Hilbert space, but the important thing is to assume there are the operators. On the algebraic approach, one can define the algebra of operators and looks for representations of the algebra on Hilbert spaces, i.e., how to realize the algebra as operators in a Hilbert space.

It turns out that for a particle with position, you still want momentum to be the generator of spatial translations. Fortunately this can be expressed as a commutation relation: this is equivalent to have operators $X,P$ with $[X,P]=i\hbar$ - this is the Canonical Commutation Relation (CCR).

Thus you look for an algebra of operators generated by $X,P$ with the relation $[X,P]$ imposed. It turns out that in the case of Quantum Mechanics (this fails just in Quantum Field Theory) the Stone-Von Neumann theorem states that up to unitary equivalence there is a single representation satisfying the CCR. They are all isomorphic to the $L^2$ representation where $X\varphi(x) = x\varphi(x)$.

So indeed you can assume there exists the Hilbert space with the operators. This justifies the abstract state space approach and justifies the existence of $P$ that acts as $-i\hbar \nabla$. Just the basis $|x\rangle$ that is non-rigorous and can only be justified in the Gel'fand triplet approach.


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.