Derivation of position operator in QM

Question

For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. 'canonical momentum' which is an operator and then apply this operator to wave functions. This is shown here in this wiki entry. We can alternatlively start with translations and use the face that the momentum operator is a generator of translations as is done here in this wiki entry.

What I am interested in is what is the more fundamental derivation for the position operator: $$\hat{X} = x.$$ To this point I have considered that the motivation for defining position operator is from the definition of the expectation value $$\langle x \rangle = \int dx x |\psi(x)|^2 = \langle \psi|x | \psi \rangle$$ where $| \psi \rangle$ is normalised. Is this the full extent of the motivation or the main point to consider? Is there another motivation for the position operator or is it taken from this motivation and confirmed by experiments to be acceptable?

Answer 1

${\hat X} = x$ because we choose to work in a basis of eigenvectors of ${\hat X}$, i.e. wave-functions.

We have a Hilbert space ${\cal H}$ which is a vector space on which we can choose any basis we wish. We most often choose to work in a basis that diagonalizes the position operator ${\hat X}$. Basis states satisfy $$ {\hat X} | x \rangle = x | x \rangle $$ Once such a basis is chosen, any state $|\Psi\rangle$ in ${\cal H}$ can be expanded in it, i.e. we can write $$ | \Psi \rangle = \int dx | x \rangle \Psi(x) $$ The function $\Psi(x)$ is called the wave-function. We can invert this to find $$ \Psi(x) = \langle x | \Psi \rangle $$

Now, what is the meaning of $\big({\hat X} \Psi \big)(x)$? By definition it is the wave-function of the state ${\hat X} | \Psi \rangle$, i.e. it is $\langle x | {\hat X} | \Psi \rangle$. It is then immediately obvious that $$ \langle x | {\hat X} | \Psi \rangle = x \langle x | \Psi \rangle = x \Psi(x) $$ Thus, we find that $\big({\hat X} \Psi \big)(x) = x \Psi(x)$. For this reason, we write for convenience ${\hat X} = x$, but one must remember that this is only true in the coordinate basis.

Momentum operator in coordinate basis: You might then ask how one can derive the expression for the momentum operator ${\hat P}$ in the coordinate basis. This is done as follows.

By definition $\big( {\hat P} \Psi \big)(x) = \langle x | {\hat P} | \Psi \rangle $. Then, $$ \big( {\hat P} \Psi \big)(x) = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | {\hat P} | \Psi \rangle = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | \Psi \rangle p = \int dx' \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | x' \rangle \langle x' |\Psi \rangle p $$ Next, using $\langle x | p \rangle = e^{\frac{i}{\hbar} p x }$, we have $$ \big( {\hat P} \Psi \big)(x) = \int dx' \frac{dp}{2\pi \hbar}e^{\frac{i}{\hbar} p (x-x') } p \Psi(x') $$ We now write $$ e^{\frac{i}{\hbar} p (x-x') } p = - i \hbar \frac{\partial}{\partial x}e^{\frac{i}{\hbar} p (x-x') } $$ Using this, we can explicitly perform the integral over $p$ and we find $$ \big( {\hat P} \Psi \big)(x) = - i \hbar \frac{\partial}{\partial x} \Psi(x) $$ Thus, in coordinate basis, we write $$ {\hat P} = - i \hbar \frac{\partial}{\partial x} $$

PS - Of course, we are free to choose any basis we want. We could for instance work in momentum basis. In this basis, we would have $$ {\hat P} = p~, \qquad {\hat X} = i \hbar \frac{\partial}{\partial p} ~. $$

Answer 2

Note that $\hat{X} = x$ is strictly not correct. The left hand side is the operator acting on the Hilbert states of quantum states. The right-hand side is a real number.

The RHS is the position-representation of the LHS. As the OP recognised correctly this is a tautology to some extent, since it is the defining representation of the operator via the eigenvalue equation

$$\hat{X} |x\rangle = x |x\rangle.$$

So the a priori definition is this one with the notion that the $|x\rangle$-states are localised delta-functions, i.e.

$$\langle x'|x\rangle = \delta^{(3)}\left(x-x'\right).$$

Note that the second equation above follows from the first equation (up to a constant) under the assumption that $\hat{X}$ is hermitian and that its spectrum is continuous.

So far this was only some maths, if you would like to read about this in a more precise manner I recommend Quantum Optics in Phase Space by W. Schleich.

So far this was only maths. The important thing which determines the physics is quantisation, which consists of imposing the relation:

$$\left[\hat{X}, \hat{P} \right] = i\hbar$$

which is also how the first relation in the question (i.e. the momentum operator in position representation) is obtained.

Answer 3

In QM you have a general postulate, that requires physical observables (momentum, angular momentum, coordinates) to be represented by Hermitian operators. Actually, in QM you have different bases as well as different notion of time evolution. Those are position and coordinate basis and Schrödinger and Heisenberg representations. Where the difference between the last two is that in the Heisenberg picture operators are time dependent. Your question is for the Schrödinger representation coordinate basis so lets concentrate on it.

In the coordinate representation, the basis is formed by an infinite set of vectors $\{| {\bf x} \rangle\}$ for which the following holds: $$ \hat{\bf X}| {\bf x} \rangle = {\bf x} | {\bf x} \rangle \,. $$
Every state vector $| \psi \rangle$ can be expressed in this basis as: $$ | \psi \rangle = \int d{\bf x} \, |{\bf x}\rangle \langle{\bf x}|\psi\rangle \,, $$ where we used the usual expression $ \int d{\bf x}|{\bf x}\rangle \langle{\bf x}|=\hat{I}$. So the position operator was easy, what about the momentum operator? Here you need more calculus. Similarly, for the momentum representation we have: $$ \hat{\bf P}| {\bf p} \rangle = {\bf p} | {\bf p} \rangle \,, $$
and $$ | \psi \rangle = \int d{\bf p} |{\bf p}\rangle \langle{\bf p}|\psi\rangle \,. $$ To relate the two we note: $$ \psi({\bf x})=\langle{\bf x}|\psi\rangle=\int d{\bf p} \langle{\bf x}|{\bf p}\rangle \langle{\bf p}|\psi\rangle=\int d{\bf p} \langle{\bf x}|{\bf p}\rangle\phi({\bf p}) \,, $$ which means that you go from one base to the other via a Fourier transformation. Where the Fourier transform is given by: $$ \psi({\bf x})=\frac{1}{(2\pi \hbar)^{\frac{2}{3}}}\int d{\bf p}\,e^{\frac{i{\bf x}\cdot{\bf p}}{\hbar}}\,\phi({\bf p}) \,. $$ In this base, for the momentum operator $\hat{\bf P}$, we have: $$ \langle {\bf x}|\hat{\bf P}|\psi\rangle = \langle{\bf x}|\hat{\bf P} \int d{\bf p}|{\bf p}\rangle \langle {\bf p} |\psi \rangle = \int d{\bf p} \langle{\bf x}|\hat{\bf P}|{\bf p}\rangle \phi({\bf p}) = \int d{\bf p} \, {\bf p}\langle{\bf x}|{\bf p}\rangle \, \phi({\bf p}) \\ = \frac{1}{(2\pi \hbar)^{\frac{2}{3}}}\int d{\bf p}\, {\bf p} \, e^{\frac{i{\bf x}\cdot{\bf p}}{\hbar}} \phi({\bf p}) \,, $$ since $$ -i\hbar {\boldsymbol \nabla} e^{\frac{i{\bf x}\cdot{\bf p}}{\hbar}} ={\bf p}\,e^{\frac{i{\bf x}\cdot{\bf p}}{\hbar}} \,, $$ we have: $$ \langle{\bf x}|\hat{\bf P}|\psi\rangle=-i\hbar {\boldsymbol \nabla} \langle {\bf x} | \psi \rangle \,. $$ Or simply, $\hat{\bf P}=-i\hbar {\boldsymbol \nabla}$, and for the other operator, $\hat{\bf X}| {\bf x} \rangle = {\bf x} | {\bf x} \rangle$, by definition, because we are working in the coordinate basis.

You can easily check that both $\hat{\bf P}$ and $\hat{\bf X}$ are hermitian and satisfy the Heisenberg relation.

P.S. You can have also the Dirac (interaction) picture where both states and operators depend on time.