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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

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 ...


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

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

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

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

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

$e^{i\theta} + e^{-i\theta}$ 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 ...


2

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: ...


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

@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

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

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 $E$.


1

It does make sense to talk about the number of field lines, but only if you take care to represent the field amplitude as being inversely proportional to the spacing between the lines. With that, the total number of lines crossing a surface is proportional to the flux, etc. Some people, notably textbook authors Chabay and Sherwood, feel that the field line ...


1

The wavefunction must be either normalizable or the limit of a sequence of normalizable functions which in general are known as distributions (generalizations of functions). A well known example of a distribution is the Dirac delta "function," $\delta(x)$. If the spatial wavefunction is $\psi=\delta(x_0)$, then the momentum wavefunction will be of the form ...


1

They have used the Schrödinger equation: $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H} \psi$$ And hence: $$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi$$ And since $1/i = -i$: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Similarly for $\psi^*$ we take the complex conjugate: $$\frac{\partial \psi^*}{\partial t} = ...


1

To arrive at the 2nd term of the third line, they have simply employed the SE equation, $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$ Re-arranging the above equation gives: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Therefore by replacing $\frac{\partial \psi}{\partial t}$ in the second line with ...


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 ...


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

No, orthogonal wavefunctions are in general not orthogonal on every subinterval. This is only true when the support of the wavefunctions (the regions in which they are nonzero) don't overlap. Energy eigenstates, for example, generally have overlapping support and are only orthogonal on whatever full interval they're defined on. To see why wavefunctions with ...


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

I know in electrical engineering, particularly in power transmission fields, sometimes people use the so-called per-unit system, where quantities are normalized to the corresponding base value. Sometimes, subscript like $U_{\text{p.u.}}=\frac{U}{U_0}$ ($U_0=U_{\text{base}}$) is used. I don't think there are any proper notations for normalized scalars. You ...


1

There is not, to my knowledge, a uniform standard on this subject. I have seen normalized quantities expressed by adding a twiddle (tilde) over the character (as in $\tilde{A}=A/A_0$), but I this notation is often reserved to indicate a time-varying quantity, instead. Unless someone knows of a strong standard in this regard, I'd say your best bet is to ...


1

This is mostly a comment but I feel it's important enough that it warrants some space as an answer. Note carefully that the wavefunction $\psi=\psi_0 \sin(kx-\omega t)$ is not a solution to the free-particle Schrödinger equation. This is easy to see mathematically. The kinetic term, $-\frac{\partial^2}{\partial x^2}$, will return a sine term, while the ...



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