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17

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) = ...


9

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 ...


8

You don't. Sinusoidal wavefunctions like these, a.k.a. plane waves, are non-normalizable, because the integral which defines the norm does not converge. $$\langle\psi|\psi\rangle = \int_{-\infty}^{\infty} |A e^{ikx} + B e^{-ikx}|^2\mathrm{d}x = \infty$$ (EDIT: just thought I should mention that $\ldots = \infty$ doesn't mean the integral literally equals ...


6

It depends on the domain of $p$. If we take the domain of $p$ to be the Schwartz space on $\mathbb{R}$, then, by symmetry of $p$, $$ \langle p^2\rangle =\langle p\psi |p\psi \rangle =\left\| p\psi \right\| ^2 $$ This is $0$ iff $p\psi =0$ iff $\psi$ is constant. However, the only constant Schwartz function is $0$. Hence, $p^2$ is positive-definite. ...


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

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 ...


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.


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

If I understand correctly, your question basically comes down to identifying a basis for the space of square-integrable functions, $L^2(\mathbb{R})$, since any physical state $|\Psi\rangle$ can be constructed by performing the integral you listed in your question with a function $\Psi_x(x)\in L^2(\mathbb{R})$. $L^2$ is known to be a vector space, so a basis ...


3

$e^{i\theta} + e^{-i\theta}$ is just $2\cos \theta$. The superposed wavefunction is $$\Psi(x,t) = 2N\cos(ax) e^{i(f(x) + \omega t)}$$ Then $$\Psi^*\Psi = 4N^2\cos^2(ax)$$ The average height is $2N^2$ if $x_0a = n\pi/2$, in which case $N = \frac{1}{2}\sqrt{1/x_0}$. Otherwise you can do this integral.


3

The eigenfunctions of a self adjoint operator lie outside the Hilbert space of square integrable functions on the line. One solution is to work with a basis of eigenfunctions of a non-self adjoint operator such as $x+ip$. Of course these are the coherent states. For the coherent states, one has an ovecomplete basis and a partition of unity, thus it is not ...


3

You test a wave function for normalizability by integrating its square magnitude. If you get a finite result then it is normalizable. To spare you complicated integrations you can also take a simpler wave function that you know is normalizable and compare it using the usual arguments. An operator is not only defined by the mathematical operation it ...


3

You can't, really. In the continuum, the "free particle wavefunction" is an abstraction to make the math easier. In a real context, you'd talk about a Wave envelope $\phi(x) = \int dk A(k)e^{ikx}$, where $A(k)$ is a bounded function that guarantees that $\int\phi^{*}\phi$ is finite and normalizable. Note that this is the generalization to the continuum of ...


3

Jerry Schirmer's answer applies in an infinite space. If you put the system in a box there is no problem: the normalized wavefunction is $\mathrm{e}^{ikx}/\sqrt{V}$. This is the usual theoretical device to make everything nice and well behaved. Then we take the limit $V\rightarrow\infty$ at the end of the day and, if we've done our job correctly, all of the ...


3

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: ...


3

There is a normalized form, though it's properly called the dimensionless Euler equations. The way to do it is define: scale time $t_0$ scale density $\rho_0$ scale length $L_0$ and then derive the scales from these: $$ v_0 = \frac{L_0}{t_0},\quad p_0=\rho_0v_0^2 $$ NB: it is possible to use other combinations, but I find that these are often the ...


3

From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense. Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be ...


3

You have the freedom to define the wavefunctions, such that one of the leading coefficients is one. Suppose you didn't do this, and instead defined: $$ \psi(x') = C_1e^{ikx'}, x' > \frac{a}{2}$$ $$ \psi(x') = C_2 e^{ikx'} + C_3 e^{-ikx'}, x' < \frac{a}{2}$$ Then, you could divide through by $C_1$ to give: $$ \psi(x') = e^{ikx'}, x' > ...


3

The wavefunction: $$ \Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)} $$ is an infinite plane wave. So it describes a particle that has an infinite extent in both time and space. That is, it exists for $-\infty \le x \le \infty$ and for $-\infty \le t \le \infty$. Unsurprisingly, if the particle has an infinite extent then it's amplitude is everywhere zero and ...


2

Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(k-k')x}\mathrm{d}x$$ show that if $\Psi(x,t)$ is normalized at time $t=0$, then the corresponding momentum space wave function $\Phi(p_x,t)$ is also normalized at time $t=0$. Good. By definition of the momentum space wave ...


2

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: ...


2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged


2

$\newcommand{\d}{\;\mathrm{d}}$You should straightforward take the integral that you mentioned: $$\int_{-\infty}^\infty \left| \psi\right|^2 \d x =\int_{-a/4}^{a/4} AA^* \d x + \int_{-\infty}^{-a/4} 0 \d x + \int_{a/4}^\infty 0 \d x =1$$ I won't do the integral and calculate $A$ because it is trivial to do therefore I leave the rest to you.


2

Let $$|n\rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}} |0\rangle$$ be the n'th normalized eigenstate and recall the commutation relation for bosonic creation/annihilation operators, $[a, a^{\dagger}] = 1$. Then we have $$a|n\rangle = \frac{ a \,(a^{\dagger})^n}{\sqrt{n!}} |0\rangle = \frac{1}{\sqrt{n!}}\big([a, (a^{\dagger})^n] - (a^{\dagger})^n ...


2

When it's written with an extra division of $<\psi|\psi>$, it is just that $|\psi>$ is not normalized.


2

Your unnormalised state is a superposition of 2 states, $\left |\uparrow \downarrow\right>$ and $\left |\downarrow \uparrow\right>$ with probability amplitudes equal to unity. Their probability amplitudes are the coefficients in front of the 2 states. In this case they are 1. But what does normalisation mean anyway? Currently as the state is, and since ...


2

The correct normalization factor is $$ N = \frac{1}{\sqrt{2}}.$$ To see this, note that you can write your wave-function in ket notation as $$\psi(x) = \langle x | a \rangle + \langle x | -a \rangle \equiv \langle x | \psi \rangle, $$ where we have used the usual basis for the (one dimensional) position representation, with normalization $$ \langle x | y ...



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