# Tag Info

8

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 $\int \psi^* \hat{O} ... 4 Normalizing$\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 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 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 ... 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 As you found, the two wavefunctions aren't orthogonal; what makes you think they should be? In general, degenerate eigenvectors of a symmetric matrix are not orthogonal. Note that since the two wavefunctions are linearly independent, the Gram–Schmidt process can be used to form an orthogonal basis for the eigenspace corresponding to the energy value$E$. 2 Both versions are wrong! The right version is, evidently, $$\langle g| g \rangle =\int g(x)^* g(x) dx = \int \left((iA f)(x) \right)^* (iAf)(x) dx = \langle iAf| iAf \rangle = \langle f| (iA)^\dagger iA f \rangle = \langle f| (-i)A^\dagger iA f \rangle = \langle f| (-i)i A^\dagger A f \rangle = \int f(x)^* (A^\dagger Af)(x) dx = \int f(x)^* (A Af)(x) dx = ... 2 As you correctly notice, this does not follow from strictly physical considerations, and it is mostly for convenience that we do this. Essentially, this simplifies quite a bit the calculations of the coefficients of your state in a given basis. Suppose, for example, that you have a basis \{\chi_+,\chi_-\} which is orthogonal but not necessarily ... 2 It is not wrong, all three normalization conditions are natural and they don't contradict each other because, in fact, the first equation is nothing else than the product of the following two equations! Just substitute your formula for \psi, \psi = R Y, to the first equation. The only mistake you have to fix to show that the first equation becomes the ... 2 @PPG: Well (−1,−1) is not a solution of your equation, but (1,−1) is... – Adam What he said was: Finding the eigenvectors You did right... but:$$(\hbar/2)\alpha+(\hbar/2)\beta=0\implies\alpha+\beta=0\implies\beta=-\alpha $$Then, the corresponding eigenvector is: c\left[\begin{array}{c} 1 \\ -1 \end{array}\right]. 2 Simple: A_0 is not a common factor to the entire wavefunction. So it's not a normalization factor. In other words, you have$$\psi(x) = A_0\sqrt{\frac{2}{L}}\sin\biggl(\frac{\pi}{L}x\biggr) + \frac{1}{2}\sqrt{\frac{2}{L}}\sin\biggl(\frac{2\pi}{L}x\biggr)$$but if you really want A_0 to be a normalizing factor you should have$$\psi(x) = ... 2 That$1/\sqrt{2}$factor is for the normalization. i.e. to ensure that$\langle 1 0 | 1 0\rangle = 1$where$\langle 1 0 |$is the conjugate transpose of$|1 0\rangle$2 From the Schroedinger equation, $$i\hbar\frac{\partial\psi}{\partial t}=\hat{H}\psi$$ simply divide by$i\hbar$: $$\frac{\partial\psi}{\partial t}=\frac{1}{i\hbar}\hat{H}\psi=-\frac{i}{\hbar}\hat{H}\psi\\$$ where we used $$\frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$ Now note that you've incorrectly applied the derivative. In quantum ... 2 The integral is a linear operation, it gets "distributed" over sums and multiplications. And given that the$\phi\$'s are the eigenfunctions, they're orthogonal to each other (they're non degenerate).

1

I think you made a mistake. Using your wavefunction and noting that $$\int^\frac{L}{2}_{-\frac{L}{2}} \sin\left(\frac{\pi x}{L}\right)\sin\left(\frac{2\pi x}{L}\right) = \frac{4L}{3\pi}$$ we see that $$1=\int^{L/2}_{-L/2}|\psi(x)|^2 dx = \frac{2}{L}\left(|A_0|^2\frac{L}{2} +\frac{4L}{3\pi} \left(A_0 + A_0^*\right)+ \frac{1}{4}\frac{L}{2}\right) = ... 1 You have to break up the domain to get rid of the absolute value. Then, do the integral by integrating by parts. 1 Any good basis should be complete. If the set of all |x> is complete, any other vector |\psi> in the Hilbert space of your system should be writable as |\psi>=\sum_{x} |x><x|\psi>. This sum does not make sense for continuous variables x, hence the need to redefine the completeness relation with an integral (as Jan's answer ... 1 I) We interpret OP's question (v2) as follows: Why not normalize$$\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 ...

Only top voted, non community-wiki answers of a minimum length are eligible