63

Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.


27

Let $$ \psi(x) = \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases} = \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) , $$ where $\mathbf 1_A$ is the characteristic function of the set $A$. Then $$ \int_{-\infty}^\infty |\psi(x)|^2 dx = \sum_{n=1}^\infty \frac 1 {n^2} < \infty, $$ ...


22

Nobody is "doing the normalization". Normalization is not even necessary. We often normalize for convenience, since that means that the Born rule for $\lvert \psi \rangle$ being the state $\lvert \phi \rangle$ reads $$ P(\psi,\phi) = \lvert \langle\psi\vert\phi\rangle \rvert ^2$$ which is certainly easier to recall/write than $$ P(\psi,\phi) = \frac{\...


19

Physicists usually generously relax the condition that the norm should be finite and they sometimes say that $|\vec r\rangle,|\vec p\rangle$ belong to the "Hilbert space". It's exactly the same "generous" language that allows physicists to say that $\delta(x)$ is a "function", the delta-function, even though its values around $x=0$ are infinite or "ill-...


17

Emilio Pisanty and Eckhard Giere have already given discontinuous, piecewise constant counterexamples in their answers. Here we provide for-the-fun-of-it a smooth infinitely-many-times-differentiable counterexample $f\in C^{\infty}(\mathbb{R})$ of a square integrable function $f:\mathbb{R} \to [0,1]$ that does not satisfy $\lim_{|x|\to \infty}f(x)=0$. Our ...


15

I) OP is right, ideologically speaking. Ideologically, OP's first eq. $$ \tag{1} \left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Turns out to be ultimately wrong!}) $$ is the statement that a particle that is initially localized at a spacetime event $(x_i,t_i)$ must with probability 100% be ...


15

Apart from not being sufficient to prove convergence of the integral $$\int |f(x)|^2\text dx<\infty,$$ having the vanishing limig $\lim_{x\rightarrow\infty}f(x)=0$ is only necessary for the convergence within a suitable class of "nice" functions. Consider, for example, the function $$ f(x)=\sum_{n=1}^\infty\chi_{\left[n,n+\frac1{n^2}\right]}(x)=\left\{\...


10

The factor of $V$ comes from integration over $\mathrm d\vec x$ (as we assume that the Hamiltonian is position independent). On the other hand the factors of $2\pi$ and $\hbar$ are essentially irrelevant: you only care about the derivatives of the logarithm of the partition function, and therefore any global factor always drops out: you can multiply the ...


10

For a quantum system with one degree of freedom on the closed interval $I$, the Hilbert space is $L^2(I)$. In this case the $I$ is the range for the spatial coordinate $x$, so that normalisation applies with respect to the Lebesgue measure on $I$. Now suppose that you have a dynamics described by the Hamiltonian $H$ on such Hilbert space, and that $\psi$ is ...


9

We know, from Schrodinger's equation, that $$ \frac{d\psi}{dt}=\left(\frac{i\hbar}{2m}\frac{d^2}{dx^2}-\frac{i}{\hbar}V(x)\right)\psi $$ Taking the conjugate tells us how $\psi^*$ evolves. $$ \frac{d\psi^*}{dt}=\left(-\frac{i\hbar}{2m}\frac{d^2}{dx^2}+\frac{i}{\hbar}V(x)\right)\psi^* $$ So let's calculate $\frac{d}{dt}|\psi|^2$. \begin{align} \frac{d}{...


9

Might as well collect my comments, most deleted, in this memo answer. Essentially, QFT does not want you to go near position eigenstates of the style of QM. The eigenstate of the momentum operator, $|p\rangle$, is not the conventional QM one, nor does it have the same dimension. However, QFT distinctly does not encourage one to seek a fantastical position ...


6

Schroedinger's equations may have both normalizable and non-normalizable solutions. The function $$ \psi_k(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2} $$ is a solution of the free-particle Schroedinger equation for any real $k$ and $\omega = |k|/c$. As a rule, if the equation has a class of solutions whose members are functions ...


6

The ambiguity is resolved once you think about the dimension of the $\delta$-function. It's actually your method of non-dimensionalization that led you astray here. Given $\delta(f(x))$, the $\delta$-function is a density with inverse dimensions of $f$. That is, $\delta(v^2 - x)$ has dimensions of inverse length while $\delta(v-\sqrt{x})$ has dimensions of ...


6

Isn't it just from the sifting property? $$f(x) = \int\mathrm{d}x'\;f(x')\,\delta(x - x')$$ That is, if you accept the above and if you accept that $$|\psi\rangle = \int \mathrm{d}x'\,\psi(x')\,|x'\rangle$$ then $$\psi(x) = \langle x|\psi\rangle = \langle x| \int \mathrm{d}x'\,\psi(x') \,|x'\rangle = \int \mathrm{d}x'\,\psi(x')\,\langle x|x'\rangle$$ $$...


6

Some simple example that illustrates that the condition $$\lim_{|x|\to \infty} f(x) = 0 \quad (1)$$ is not necessary. If the condition were necessary $f\in L^2$ would imply that the limit in (1) holds. Take in dimension 1 the function $$ f(x) = \sum_{n=2}^{\infty} \chi_{I_n}(x) $$ where $\chi_{I_n}$ is the characteristic function of the interval $I_n = [...


6

This is true not just for position eigenvectors. This is true for all eigenvectors in their own (orthonormal) eigenbasis. For an operator with discrete eigenvalues $n$ with eigenvectors $|n\rangle$, $$\langle n'|n\rangle=\delta_{n,n'}$$ where $\delta_{n,n'}$ is the Kronecker Delta that is $1$ when $n=n'$ and $0$ otherwise. For an operator with a ...


5

First, if you want to normalize it the wave function needs to have some free constant, so $\psi(x)=A\,(\phi_1(x)+2\phi_2(x)+3\phi_3(x))$. Then normalize as you said: $$\int\left|\psi(x)\right|^2dx = A^2\;(\int\left|\phi_1(x)\right|^2dx + 4\int\left|\phi_2(x)\right|^2dx + 9\int\left|\phi_3(x)\right|^2dx + \text{cross terms})=1$$ The eigenfunctions of a ...


5

This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...


5

It doesn't matter what sign you choose. Notice that since $|A|^2 = \frac{2}{a}$, you could even pick $A = \sqrt{\frac{2}{a}} e^{i\phi}$, so $A$ doesn't have to be real. The reason is that a wavefunction is only defined up to a global phase. The reason is that we calculate probabilites with $|\psi|^2$ and mean values of operators with $\int \psi^* \hat{O} \...


5

$\newcommand{\ket}[1]{\left|#1\right\rangle}$ $\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\braket}[1]{\left\langle#1\right\rangle}$ After reading from Yeh's notes on Advanced Condensed Matter Theory, Section II.9, I've decided that Kleinert was right about the numerator, but the denominator actually comes from normalization. So, you start ...


5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...


5

In Quantum Mechanics, a physical state is always normalized by definition. So, in fact, you have to normalize your linearly dependent vectors, obtaining the same normalized vector, and, in fact, both represent the same physical state. Norm is not important. Note however that scalars in QM are complex numbers, so your two vectors could be $|{\psi_1} \rangle$ ...


5

The formula for the expectation value $\langle A\rangle=\langle\psi|\hat{A}|\psi\rangle$ is given for the normalized states $\langle\psi|\psi\rangle=1$. You can generalize it as \begin{equation} \langle A\rangle=\frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle} \end{equation} Of course this expression would still be ill-defined for $|x\rangle$ ...


5

The set ${\cal L}^2(\mathbb{R}^3)$ of square integrable functions $\psi:\mathbb{R}^3\to \mathbb{C}$ is a $\mathbb{C}$-vector space, and hence includes $0$. The set $\{\psi\in{\cal L}^2(\mathbb{R}^3)\mid ||\psi ||=1\}$ of normalized wavefunctions is not a vector space, e.g. because the set is not stable against multiplication with a constant. The set $\{\psi\...


4

Normalizing $\psi$ to $1$ means that we ensure that $$ \int|\psi|^2dx = 1 $$ normalizing it to $-i$ would presumably mean ensuring that $$ \int|\psi|^2dx = -i $$ which is impossible because the integrand $|\psi|^2$ is positive everywhere.


4

The physical interpretation of the wavefunction is that it's amplitude squared tells you the probability of finding the particle described by that wavefunction at a certain location: $$P(\text{particle at $x$}) dx = |\psi(x)|^2 dx$$ The probability to find it in some interval is then given by the integral of the amplitude squared over that interval: $$P(\...


4

I) We interpret OP's question (v2) as follows: Why not normalize $$\tag{1} \langle x_1 | x_2\rangle~=~\delta_{x_1,x_2}~:=~\left\{ \begin{array}{ccl} 1 & \text{for} & x_1= x_2, \\ 0& \text{for} & x_1\neq x_2, \end{array} \right. $$ via a continuous Kronecker delta function rather than a Dirac delta distribution $$\tag{2} \langle ...


4

Suppose $\psi$ satisfies the (dimensionless) time-dependent Schrödinger equation: $$ i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi. $$ It will also satisfy the conjugate equation: $$ -i\frac{\partial\psi^*}{\partial t}=-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^* $$ Now consider how the normalisation changes over time: $$...


4

We divide by the norm of $\left|\psi\right>$ in order to take into account the case where the vector is not normalised. If $\langle \psi|\psi\rangle=1$ it makes no difference and if it's not, you recover the correct result as well.


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