Applying measurement postulate to a continuous sum of eigenvectors (by analogy)

Question

Measurement postulate:

If we measure the Hermitian operator $\hat Q$ in the state $Ψ$, the possible outcomes for the measurement are the eigenvalues $q_1$, $q_2$, . . .. The probability $p_i$ to measure $q_i$ is given by $$p_i = |α_i |^2 $$ where $Ψ(x) = \Sigma α_iψ_i(x)$

Trying to apply it to a continuous sum of eigenstates:

I am trying to apply the measurement postulate to a continuous sum of eigenstates.

In the continuous sum $$ \Psi(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{ikx}dk$$

• $e^{ik_0x}$ plays the role of the eigenstate $ψ_i(x)$, for a particular $k_0$

• $\frac{\Phi (k_0) dk_0}{\sqrt{2\pi}}$ plays the role of the coefficient $\alpha_i$

• the operator $\hat Q$, for example, is $\frac{\hbar}{i} \frac{\partial}{\partial x} $

• the eigenvalue associated to $e^{ik_0 x}$ is $p_0$

As a result of this analogy, it should follow that the probability to measure $p_0$ in the state $Ψ$ is: $$ \frac{|\Phi (k_0)|^2 |dk_0|^2}{{2\pi}} $$

I know however it is not correct because:

• of this weird $|dk_0|^2$,

• I know that $\int |\Phi(k)|^2dk = 1$ (from Parseval theorem) so the $2\pi$ looks suspicious

Edit:

I realize that if I've had taken the other Fourier transform (the one with the $2\pi$ in the exponential), the problem with the $2\pi$ would not have arised!

Answer 1

I've changed your bullet points: In the continuous sum $$ \Psi(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{ikx}dk\,,$$

• $\frac{e^{ik_0x}}{\sqrt{2\pi}}$ plays the role of the eigenstate $ψ_i(x)$ for a particular $k_0$, rather than just ${e^{ik_0x}}$, because the state needs to be normalized (in a sense to be discussed below)

• ${\Phi (k_0)}$ plays the role of the coefficient $\alpha_i$; we don't include the $dk$ for reasons explained below, and we've already moved off the factor of $\sqrt{2\pi}$

• etc.

In moving from the discrete case to the continuous case, we can no longer talk about the probability of a getting a particular value upon measuring an observable. Instead, we proceed as follows.

First of all, imagine we're measuring a continuous variable, say the position of a particle, but we can only measure the position to be in certain regions rather than at particular places (e.g., we lack perfect precision in our measurement. In other words, we have bins of size, say, $\Delta x$, so that we measure the position to be in $\dots,[-2\Delta x, -\Delta x]$, $[-\Delta x, 0]$, $[0, \Delta x]$, $[\Delta x, 2\Delta x],\dots$ and so on. Then, we can talk about the probability $P_j$ of measuring the position to be in bin $j$, i.e., in the bin $[j\Delta x, (j+1)\Delta x]$ so that the probability of finding the position being anywhere is of course 1, i.e. $$ 1= \sum_{j=-\infty}^{\infty} P_j\,. $$ Now, imagine that we want to make $\Delta x$ smaller and smaller, so that we are refining the precision of our measurement. In fact, let's take the limit as $\Delta x\to0$. How do we do this? Well, write $$ 1= \sum_{j=-\infty}^{\infty} {P_j} =\sum_{j=-\infty}^{\infty} \frac{P\left(j\Delta x\leq x \leq(j+1)\Delta x\right)}{\Delta x}\Delta x\,. $$ In order for the limit to make sense, the quantity $P_j/\Delta x$ should approach a finite limit $p(x)$ as $\Delta x \to0$. We can then interpret $p(x)$ as a probability per unit length; that is, it is a probability density. The assumption that the limit exists is a reasonable assumption, because it essentially means that as long as $\Delta x$ is small enough, doubling the interval implies that we've doubled the probability that we can find the particle in the interval.

Now, accordingly, $p(x)\Delta x$ is approximately the probability that the particle is found between $x$ and $x+\Delta x$, and hence the probability of finding the particle at $x$ is zero. This is just the way that probability distributions for continuous variables work. Finally, then, the sum above becomes the limit, i.e. $$ 1=\int_{-\infty}^{\infty} p(x)\,dx\,. $$

Moving back to the quantum discussion, then, $|\Phi(k)|^2$ plays the role of a probability density function, and therefore $|\Phi(k)|^2dk$ is the probability that the momentum of the particle is measured to be in the interval $[k, k+dx]$. We can verify that this makes sense at least mathematically buy using the (distributional) fact that $$ \delta (k-k_0) = \int_{-\infty}^{\infty}\frac{e^{i(k-k_0)x}}{2\pi}dk $$ to verify that $$ 1 = \int_{-\infty}^{\infty} |\Phi(k)|^2dk\,. $$ (This is why $\sqrt{2\pi}$ is attached to the basis function: it is essentially the normalization factor of the basis function. Re the last comment in the OP about the factor of $2\pi$ in the exponent of the exponential: this would indeed render the normalization factors equal to 1. However, note that $$ \frac{\hbar}{i} \frac{d}{dx}e^{i2\pi k x} = (\hbar 2\pi k) e^{i2\pi k x}\,, $$ so that the eigenvalue of the momentum operator is $p = \hbar 2\pi k$. This means that you have to re-interpret the value $k$: it is no longer the wave number; instead it is the reciprocal of the de Broglie wavelength, directly. This is fine, but it's not the convention normally chosen.)

Answer 2

During the transition from finite to continuous basis $\nu_k$ of eigenstates we can't save the equation: $$|\Psi\rangle =\sum|\nu_k\rangle \Psi_k$$ where $\Psi_k=\langle \nu_k|\Psi\rangle ,$ because transition is described by limit (usually we just split the part of value axis to finite number (N) of possible eigenvalues, coordinates for ex.): $$N\rightarrow \infty$$ The consequences of this will be the infinite sum $$\Psi=\sum |\nu_k\rangle \Psi_k$$ with senseless infinite value. The solution of this problem is increasing the eigenvector normalization $$||\nu_k\rangle |^2= 1\rightarrow ||\nu_k\rangle |^2= 1/\epsilon$$ where $\epsilon$ is the minimal distance between allowed eigenvalues ($q_{k+1}-q_{k}=\Delta q$). It will cost us impossibility to realize the eigenstates in real life, because they can not be normalized to 1 in real life, that's why, for example the diffraction phenomena exists and photon wave package can't consist of only one clear monochromatic wave.

It's easier to write normalization by conjugate vectors: $$|\nu_k\rangle ^{T*}=|\nu_k\rangle ^{+}=\langle \nu_k|\quad \text{so called bra-vector}$$ So, in general it's usually written like: $$\langle \nu_n|\nu_m\rangle =\dfrac{1}{\epsilon}\delta_{nm}$$ So, we have to make transition because of this renormalisation: $$\Psi=\sum_{k=1}^N|\nu_k\rangle \Psi_k=\sum_{k=1}^N|\nu_k\rangle \langle \nu_k|\Psi\rangle \rightarrow \Psi=\sum_{k=1}^N|\nu_k\rangle \dfrac{\langle \nu_k|\Psi\rangle }{\langle \nu_k|\nu_k\rangle }=$$ $$=\sum_{k=1}^N|\nu_k\rangle \epsilon\langle \nu_k|\Psi\rangle $$ In the limit we easily obtain: $$\lim_{N\rightarrow \infty} \sum_{k=1}^N|\nu_k\rangle \epsilon\langle \nu_k|\Psi\rangle =\int |q\rangle dq \Psi(q)$$ In your cases we can choose the impulse basis: $$|\Psi\rangle =\int |p\rangle dp \Psi(p)\rightarrow \langle q|\Psi\rangle =\Psi(q)=\int \langle q|p\rangle dp\Psi(p)=\int \Psi_p(q)dp\Psi(p)$$ The probability of measure $p_0$ is still $\Psi(p_0)$ but you can't anymore use $$|\Psi\rangle =\sum|\nu_k\rangle \Psi_k$$