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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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