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## New answers tagged wavefunction

0

There are two ways to interpret the boundary conditions you are imposing. The first case is that of a system which is infinite in extent, but has a periodic regularity. This is like an electron in an idealised 1D crystal, where the periodic boundary condition is imposed by the presence of nuclei regularly spaced. In this case, the plane wave solution has ...

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This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land ... 8 Notice that$\psi(x)$is defined on a circle of circumference$a$. Multiplying$x$on this circle is really multiplying a periodic extension of$x$, i.e., the sawtooth function$x - a\lfloor x/a\rfloor$, where$\lfloor y\rfloor$means the largest integer not greater than$y$. So, the commutator of the position and momentum operators involves the derivative ... 0 Energy or the value of$V(x)$negative means it is a bound system. Think of it in this way, if a particle is free and has no kinetic energy and potential energy then it's total energy is zero. If this particle is not free or otherwise is bounded by a negative potential well then it's potential energy is$-V$. You have to give the same amount of energy, in ... 3 No. The$p_x$and$p_y$orbitals are always real-valued, and the complex-valued$|m|=1$orbitals are always denoted$p_{1}$and$p_{-1}$. Depending on what scheme you're using, the third$l=1$orbital can be denoted either$p_z$or$p_0$, with both notations completely equivalent. Just because the$p_x$and$p_y$orbitals are linear combinations of$p_1$... 2 You have stumbled upon a difference between how chemists and physicist denote orbitals. The$p_x, p_y, p_z$notation is common in chemistry because the resulting orbitals are real. Physicist use$p_{-1}, p_0, p_1$where the subscripts are the values of m and embrace the complexity of the resulting wave function. It is purely a matter of choice since all ... 1 So to your first question, when you calculate the time-evolution of a wavepacket, you first transform your wavepacket representation from a positional basis to a momentum basis. $$|\psi\rangle = \sum_k c_{x,p}|p\rangle$$ This moment basis is useful as they are eigenstates to your hamiltonian and satisfy $$H |p\rangle = E_p |p\rangle$$ This in turn allows ... 1 Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2. 0 Such a vector describes the quantum state of a spin-1 particle on a line, or any other particle with a position degree of freedom and 3 internal states. To start with, you can expand wavefunctions in a basis. E.g., if you have a wavefunction$\vert\psi\rangle$which depends on position, you can expand it in the position basis$\vert x \rangle_p$, i.e., ... 1 In general, what we use for periodic boundary conditions is defined by$Y(0)=Y(L)$and$\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition$Y(0) = Y(L)$I believe is not sufficient to impose conditions on$k$. This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the ... 0 For the strict boundary conditions, the -n solution is linearly dependent with the n solution: it's the same wavefunction multiplied by -1.$\psi_{-n}(x)=\sin(-n\frac\pi Lx) = -\sin(n\frac\pi Lx)=-\psi_{n}(x)$2 Yes, you multiply each of the super-positioned states (energy eigenstates) with their time evolution as $$\Phi(x,t) = \phi_n(x)e^{i E_n t/\hbar} + \phi_{n+1}(x)e^{i E_{n+1} t/\hbar}$$ here$\phi_n(x)$is the first state and$\phi_{n+1}(x)\$ is the second. As for the frequency of this state, try to write it as $$\Phi(x,t) = e^{i \omega t} ... 3 It's not "multiplied by r^2 to get the probility density". The issue is that the volume element in spherical coordinates is$$ \mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$and since the probability to find a particle in a subspace X\subset \mathbb{R}^3 is$$ P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$by definition of a ... 2 Note: All products between \chi's are to be understood as tensor products. Assuming the fermions are both spin 1/2 particles, one recalls that the spin-part of the total wavefunction in the spin 1 triplet is either one or a linear combination of$$ \chi(j=1,m=1) = \chi(1/2,1/2)\chi(1/2,1/2)  \chi(j=1,0) = \frac{1}{\sqrt{2}}\big( ...

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You cannot measure the energy of a single particle constituent of a system of N interacting (bound) particles. It is entangled with all the other constituents and you cannot measure (disturb) one of them without changing the others. You can measure its removal energy but this is not an energy associated with it alone. Theoretically you can come close to ...

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