# Representations

To solve the Schrödinger equation for a given problem explicitly we need to choose a representation. For a free particle the Hamiltonian is \begin{align} H=\frac{p^2}{2m} \end{align} in Quantum Mechanics $H$ and $p$ become operators. This is sometimes indicated by a hat \begin{align} H=\frac{p^2}{2m} \rightarrow \hat{H}=\frac{ \hat{p}^2}{2m} \end{align} In a similar fashion the Poisson bracket have an analog in Quantum Mechanics \begin{align} \{ x, p \} \rightarrow [ \hat{x},\hat{p} ] \end{align} they become the commutator $[\, , \, ]$ of the operators $\hat{x}$ and $\hat{p}$, which play a fundamental role in the uncertainty principle, as was illustrated at Heisenberg's Microscope.

The Schrödinger equation for a free particle becomes \begin{align} i \, \hbar \frac{d}{dt} |\psi(t) \rangle & = H \, |\psi(t) \rangle = \frac{\hat{p}^2}{2m}\, |\psi(t) \rangle \end{align} to solve this equation we represent the operators and the state vector in an appropriate basis. A convenient choice for a basis is one in which eigenstates of the operator correspond to a physical quantity, namely the position and the momentum. Thus we will now discuss the position or the momentum basis.

## Position Basis

We choose a basis $\{ |\phi_x \rangle \}$ for which the $|\phi_x \rangle$'s are the eigenvectors of the operator $\hat{x}$. For a particluar eigenvector $|\phi_{x_0} \rangle$ the position $x_0$ is the corresponding eigenvalue. $|\phi_{x_0} \rangle$ satisfies the eigenvalue equation \begin{align} \hat{x} | \phi_{x_0} \rangle =x_0 | \phi_{x_0} \rangle \end{align} we can condense the notation by simply writing \begin{align} \hat{x} | {x_0} \rangle =x_0 | {x_0} \rangle \end{align} where the position $x \in \mathbb{R}$ can take any real value. Thus we have a continuous basis for which the orthogonality relation becomes \begin{align} \langle x_i | x_j \rangle=\delta_{ij} \quad \rightarrow \quad \langle x | x' \rangle= \delta(x-x') \end{align} the compleness relation changes to the integral \begin{align} \sum_i | x_i \rangle \langle x_i|= \mathbb{I} \quad \rightarrow \quad \int dx \, | x \rangle \langle x| = \mathbb{I} \end{align}

The state vector can be represented in the position basis by the expansion \begin{align} |\psi (t) \rangle = \sum_i c_i \, |x_i \rangle \quad \rightarrow \quad |\psi (t) \rangle = \int dx \, c_x(t) \, |x \rangle \end{align} and by completeness \begin{align} |\psi (t) \rangle &= \mathbb{I} \, |\psi (t) \rangle \\ &= \int dx \, | x \rangle \langle x|\psi (t) \rangle \\ &= \int dx \, \langle x|\psi (t) \rangle \, | x \rangle \\ \end{align}

Thus the probability amplitude is given by \begin{align} c_x(t)= \langle x|\psi (t) \rangle \end{align} We use a more convenient notation and write the probability amplitude as psi, depending on the continuous position $x$ and time $t$ \begin{align} c_x(t)= \langle x|\psi (t) \rangle =\psi(x,t) \end{align} which is the position space wave function.

## Momentum Basis

The representation in the momentum basis $\{ |\phi_p \rangle \}$ or just $\{ |p \rangle \}$ can be derived in a similar fashion as the position basis. Here the eigenvector $|p_0 \rangle$ satisfies the eigenvalue equation \begin{align} \hat{p} |{p_0} \rangle =p_0 |{p_0} \rangle \end{align} The orthogonality and completeness relation are true as well \begin{align} \langle p | p' \rangle= \delta(p-p') \end{align} \begin{align} \int dp \, | p \rangle \langle p| = \mathbb{I} \end{align} expanding $|\psi (t) \rangle$ in the momentum basis \begin{align} |\psi (t) \rangle &= \mathbb{I} \, |\psi (t) \rangle \\ &= \int dp \, | p \rangle \langle p|\psi (t) \rangle \\ &= \int dp \, \langle p|\psi (t) \rangle \, | p \rangle \\ \end{align} we find the probability amplitude in the momentum basis \begin{align} c_p(t)= \langle p|\psi (t) \rangle \end{align} and call it \begin{align} c_x(t)= \langle p|\psi (t) \rangle =\tilde{\psi}(p,t) \end{align} the momentum space function.

• Given a physical system in the momentum eigenstate $|p_0 \rangle$ what's the probability amplitude of being at the position $|x \rangle$?
• Are these observerables simultaniously measureable?
• Can $|x_0 \rangle$ be an eigenstate of the operator $\hat{p}$?

In another article we discussed that the change of basis between $\tilde{\psi}(p,t)$ and $\psi (t)$ are related by the (inverse) fourier transform. We found that the probability amplitude, that when the position of the particle is $x$ the momentum $p$ is represented by the scalar product$\langle p |x \rangle=e^{i \, p \cdot x / \hbar}$. The uncertainty principle states, that canonically conjuagete variables can't be mearured arbitrariliy accurate. The more precise we know the momentum, the less accurate we can determine the posision and vice versa. If we know the position exactly, the probability amplitude in the momentum space is uniformly distributed. This is also reflected by the fact that the fourier transform of the delta distribution is a constant.

The last question is of special interest since we often work in the position basis to solve Schrödinger's equation \begin{align} i \, \hbar \frac{d}{dt} \langle x |\psi(t) \rangle & = \langle x | \left( \frac{\hat{p}^2}{2m} + V(\hat{x}) \right) |\psi(t) \rangle \end{align}

this leads us to the question, wheather or not the operators $\hat{x}$ and $\hat{p}$ commute. It can be shown, that

• if two operators $A$ and $B$ commute $[ A,B ]=0$, then they have the same set of eigenvectors and are simultaniously diagonalisable.
• if they don't commute $[ A,B ] \not=0$ then they cannot be simultaniously diagonolised, but they may share some eigenstates.

## Mean Values in the Position Basis

As introduced with the Postulates the mean value of an observable is \begin{align} \langle A \rangle= \langle \psi | A | \psi \rangle \end{align} in the position basis $| \psi \rangle$ becomes \begin{align} \langle x | \psi \rangle = \int dx \, \psi (x,t) |x \rangle \end{align} and \begin{align} \langle \psi | x' \rangle = \int dx' \, \langle x' | \, \psi^* (x',t) \end{align} Thus the mean of the operator $A$ in the position basis is \begin{align} \langle A \rangle= \int dx \int dx' \psi (x,t)\psi^* (x,t)\langle x' | A | x \rangle =\int dx \int dx' \psi (x,t)\psi^* (x',t) A(x,x') \end{align}

## Representation of Operators

Finally we try to find out what $\hat{p} | x \rangle$ is in the position basis. To do this we need to represent the operator $\hat{p}$ in the position basis. We start the argument from the uncertainty principle $\hat{x}\hat{p}-\hat{p}\hat{x}= i \hbar \mathbb{I}$ in the position basis \begin{align} \langle x|\hat{x}\hat{p}-\hat{p}\hat{x}|x' \rangle \end{align} $\hat{x}$ acting on $|x' \rangle$ gives the eigenvalue $x'$ similarly $\hat{x}$ acting on $\langle x|$ gives the eigenvalue $x$ \begin{align} \langle x|\hat{x}\hat{p}-\hat{p}\hat{x}|x' \rangle=x \langle x|\hat{p}|x' \rangle - x'\langle x|\hat{p}|x' \rangle= (x-x') \langle x|\hat{p}|x' \rangle \end{align} by the uncertainty principle this is equal to \begin{align} (x-x') \langle x|\hat{p}|x' \rangle = i \hbar \langle x|\mathbb{I}|x' \rangle= i \hbar \langle x|x' \rangle = i \hbar \delta(x-x') \end{align} now we have an expression for the operator in the position basis \begin{align} \langle x|\hat{p}|x' \rangle = i \hbar \frac{\delta(x-x')}{(x-x')} \end{align} Now we claim, that $\frac{\delta(x-x')}{(x-x')}$ is just the derivative of $-\delta(x-x')$ (watch the video lecture for an intuitive but non-rigorous explanation). The operator $\hat{p}$ acting on $|x' \rangle$ in the position basis is thus given by \begin{align} \langle x|\hat{p}|x' \rangle = -i \hbar \frac{\partial}{\partial x} \delta(x-x') \end{align} By the superposition principle $\hat{p}$ acting on any state vector $|\psi \rangle$ is \begin{align} \langle x|\hat{p}|\psi (t) \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x,t) \end{align} Therefore we associate

the momentum operator $\hat{p}$ in the position basis with the derivative \begin{align} \hat{p}=-i \hbar \frac{\partial}{\partial x} \end{align}

Similarly we can find out what $\hat{x} | p \rangle$ is in the momentum basis and find that

the position operator $\hat{x}$ in the momentum basis \begin{align} \hat{x}=i \hbar \frac{\partial}{\partial p} \end{align}

Video Lectures:

• V. Balakrishnan - Quantum Physics Lec 7 [1]