New answers tagged

3

It is because the first excited level is the lowest energy state that is orthogonal to the ground state. The ground state is symmetric under $x\to-x$ and the first excited state is antisymmetric. So, if we explore the energy-expectations of state that are antisymmetric, the trial states are orthogonal to the ground state and the lowest energy one will ...


0

Mirror harmonics of the hydrogen wave function. Resonant frequency $1.42\ GHz$


2

You can define $|\Psi\rangle$ as: \begin{equation} |\Psi\rangle=\int \Psi(y) |y\rangle d^3y \end{equation} With $\{|y\rangle \ \,|\,y \in \mathbb{R}^3\}$ the basis of the hilbert space $H$ of positions. Since $\langle x|$ is the linear form such that $\langle x|(|y\rangle)\equiv \langle x|y\rangle=\delta^{(3)}(x-y)$ we have : \begin{equation} \langle x|\Psi\...


12

Consider the case of a vector space of countable dimension, with some orthonormal set of basis kets $\left\{\vert\mathbf{e}_i\rangle\right\}$. The orthonormality condition is stated as $\langle \mathbf{e}_i \vert \mathbf{e}_j \rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. We can then expand any vector in this basis, $$\vert \psi \rangle =...


1

I don't think this is the most general answer, but it is an answer. In the case of translations, let your orthogonal wavefunctions be eigenstates of the momentum operator, which generates translation. For example, take $\phi_{p_1}(x)$ and $\psi_{p_2}(x)$ to be two mutually orthogonal eigenstates of the momentum operator with eigenvalues $p_1$ and $p_2$, ...


1

If the Schrödinger equation yields solutions that show the probability as the square of the amplitude, then the ‘negative’ Schrödinger equation’s solution is an operator √(1−x2) applied to the normal solution. This expression doesn't make sense on dimensional grounds. If you intend $x$ to be the wavefunction, then $x^2$ has units, so you can't subtract it ...


3

In essence, the answer is yes. You do it by going to the "energy-space", i.e., you use the energy eigenbasis to span your Hilbert space. Let's say the kets forming the energy eigenbasis are $\vert E \rangle$ then the wavefunction of a state $\vert \psi\rangle$ in this basis would be given by $\psi(E) \equiv \langle E \vert \psi \rangle$. Just like $...


3

The basic idea of the Fourier transform is that of a transformation between two different bases in a vector space: In the position representation, the basis is that of states with well-defined positions $\phi_x$ (i.e. Dirac delta functions), and the state is written as $\psi = \int \psi(x)\phi_x \mathrm dx$, i.e. as a linear combination of the basis ...


0

We can write $$\psi_{k + \frac{2\pi}{a}}(x) = e^{i(k + \frac{2\pi}{a})x}u(x) = e^{ikx}e^{2\pi i\frac{x}{a}}u(x)$$ But $e^{2\pi i\frac{x}{a}}$ has period $a$, so the function $$u'(x) \equiv e^{2\pi i\frac{x}{a}}u(x)$$ is also periodic with period $a$. Our original wave is thus equivalent to $$ \psi'_k(x) = e^{ikx}u'(x)$$ which is a Bloch wave at wavevector $k$...


0

Another way to deduce which direction the wave is traveling is to see when the phase of the wave remains constant as time evolves. Your time-dependent wavefunction would be $$\Psi(x)=Ae^{-iwt}e^{ikx}+Be^{-iwt}e^{-ikx}$$ $$=Ae^{-i(wt-kx)}+Be^{-i(wt+kx)}$$ So if time increases, the phase of the first term will be constant if x increases. This means the plane ...


0

Consider a wave function F(x) = Asin($\frac{2\pi}{λ}x$) now differentiating on both sides we get F'(x) = $\frac{2\pi}{λ}Acos(\frac{2\pi}{λ}x$) differentiating again, we get F''(x) = -$\frac{4\pi^{2}}{λ^{2}}Asin(\frac{2\pi}{λ}x$) ∴ F''(x) = -$\frac{4\pi^{2}}{λ^{2}}F(x)$ for a particle like electron, moving in 3D (arbitrary) manner, we must consider x, y, z ...


1

Fields in quantum mechanics are not measurable quantities, and also wavefunctions are not measurable. What are measurable are interactions between elementary particles or composites of elementary particles, like atoms and nuclei. Fields and wavefunctions are mathematical representations that allow a predictive modeling of these interactions . The predictions ...


1

then it will take some time for the wave function of particle B to collapse after the measurement of A This is your problem. It doesn't take any time- wave function collapse is instantaneous, in every reference frame. That is to say, once a measurement has been taken, all further measurements are taken on the collapsed wavefunction. There is no violation of ...


1

A general free-particle state takes the form $$\psi(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \phi(k) \exp[ikx]dk$$ The average momentum of this state can be shown to be $$\left<\hat p \right> = \frac{1}{2\pi}\int_{-\infty}^\infty \hbar k |\phi(k)|^2 dk$$ If the function $|\phi(k)|^2$ is even, then $\left<\hat p\right>=0$; of course, this need not ...


2

$$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int\frac{d}{dt}\left(\psi^*\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By Leibniz's Rule}$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int \left(\frac{d\...


0

A preliminar point should be made clear, before discussing Uncertainty Relations. According to Quantum Mechanics a particle is never a wave or a superposition of waves. Classical waves have nothing to do with individual quantum particles. A relation with waves does exist, but it is much more subtle: waves can used to obtain the probability distribution of ...


0

You more or less answered your own question. The wave properties are described by a wave function that is governed by the dynamical equations of the system under investigation and the statistical properties appear when you make measurements of this wave function. Such measurements will sample the wave function in a random fashion so that the squared modulus ...


2

You can use the projection operators $$ P_\pm = \frac 12 (1+{\bf n}\cdot {\boldsymbol \sigma}). $$ Applied to any starting state they give you the eigenstates of ${\bf n}\cdot {\boldsymbol \sigma}$ with spin $\pm$ along the direction specified by the unt vector ${\bf n}$. For small matrices projection operators are usually the fastest route the eigenvectors.


1

To compute the spin components along $\hat{n}$ consider the matrix $$ \sigma_{n} \equiv \hat{n} \cdot \vec{\sigma}\,. $$ Diagonalize $\sigma_n$ and find the list of eigenvectors $|i\rangle_n$. Then express your ket in terms of that basis. So find the coefficients $c_i$ in $$ |\mathrm{your\;state}\rangle = \sum_i c_i \,|i\rangle_n\,. $$ The probability of ...


0

Is it possible that during bonding, these electrons still behave as waves and interfere? Of course. This has been confirmed experimentally. The unequivocal experimental proof of the wave (frequency) nature of matter is the electron image recorded in 2007. This image was recorded by a team of Swedish scientists from the University of Lund. The image turned ...


4

The wavefunction you are giving is the one of a particle (with no spin) in an infinite potential well, this is described as a state living on a certain Hilbert space. To include the spin of a particle you must force it as a tensor product of the system you are considering times the two-level system from the spin (if you want to consider the potential well $\...


1

imagine that you have two particles that first interact in such a way that their combined spin wave function becomes a superposition of one having spin up and the other spin down, and vice versa. Spin and magnetic dipole of subatomic particle are correlated. The production of superposition means that these two magnetic dipoles are oriented anti-parallel ...


4

You are asking the right kind of question, and the main thing I want to say is to warn you that a huge amount of stuff has been written on this and only about $0.00001$ percent of all that stuff is worth reading. These questions go to the heart of what we mean by a 'state' when talking about quantum systems. Most people try to frame their discussion by ...


4

This arises from a fundamental misconception in the early days of quantum mechanics, that the quantum state describes the physical state of a particle. In fact the quantum state describes an observer's knowledge of the particle (well illustrated in Schrodinger's cat and Wigner's friend). So, if Alice measures particle A, she acquires knowledge of particle B, ...


1

Make your measurement at B. You'll get either "up" or "down". Repeat the whole experiment many times. You'll get "up" about half the time. Is this because the wave function is collapsing as you measure, or because it's collapsing shortly before you measure? There's no way to tell. Either description leads to the observed 50/...


0

the concept of what happens before loses meaning when the events have a space-like separation. In fact geometrically speaking, this whole business of saying before for such events is ill-defined. There is no strict time ordering possible between both events precisely because any ordering is observer-dependent What do they really mean when you say that in ...


1

This is not the way to solve two simultaneous coupled differential equations. (To be honest, I'm not completely sure how exactly you managed to get what you say you did!) I won't solve the whole problem, but I'll give you a couple of pointers of how to get to the solution. The point is that rate of change of the variables $c_x$ depends not only on the value ...


0

If your incident wavefunction is $$\psi_i=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty{A(k)e^{i(kx-\omega_kt)}dk},$$ the transmitted wavefunction is $$\psi_t=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty{\alpha(k)A(k)e^{i(kx-\omega_kt)}dk}$$ where the transmission coefficient for an incident wave $e^{ikx}$ is $T(k)=|\alpha(k)|^2$. The transmission ...


0

Related question, but I'm sure there are many answers out there, so I won't say much. I think the reason is basically that it makes sense and it works. Formally, I think it is generally considered as a postulate which makes sense only because experiments agree with it.


1

This is perhaps the deepest mystery in understanding quantum mechanics, and the answer is generally only explained in postgraduate level mathematics. We can start by observing that there is randomness in almost any experimental result. Classically, this is understood in terms of experimental accuracy, but it is also a general principle on which we can base ...


2

The only justifications I've seen were always the same: ... [De Brogile]...[Planck]...[Dispersion] But thats it. Thats the ultimate reason. There isn't any deeper explanation. You can mathematically revolve around in a lot of ways but this doesn't explain it on a deeper level whatsoever. So let's look at your actual question here: But a plane wave is only ...


3

Another way to look at it (but not necessarily more "fundamental") is to see that the translation operator of a localised particle $\left|x\right>$ can be expressed as a Taylor series operator: $$\left|x-a\right> = e^{a\frac{d}{dx}}\left|x\right>$$ Since translations have to leave the momentum invariant, we need to require that the ...


10

First, an important note - $i\hbar\frac{\partial}{\partial t}$ is not the energy operator. It is not an operator at all. Remember that an operator acts on elements of the Hilbert space, such as $L^2(\mathbb R)$; time derivatives do not act on these functions. See my answer here. The most fundamental justification I know has to do with symmetry groups. To ...


3

The justification is heuristic. Start with the plane wave: $$ \Psi(x,t)=e^{i(kx-\omega t)} $$ The momentum $p=\hbar k$ “is recovered” by taking $-i\hbar \frac{\partial\Psi(x,t)}{\partial x}$ and the energy “is recovered” by taking $i\hbar\frac{\partial\Psi(x,t)}{\partial t}$. Thus, the energy relation for a free particle, described by a plane wave, is $$ E=\...


3

So let's start without spin. You can extract the wavefunction from the 'ket vector' by taking the inner product with the $|x\rangle$ state. The $|x\rangle$ ket represents a state with definite position where a particle is localised entirely on $x$. This is not a physical state (you can't normalise it) but still a useful tool. The wavefunction is then ...


5

I will restrict my answer to the 1-dimensional momentum operator, which is enough to understand what is going on. The momentum operator you have written has the following form in 1D: $$ \hat{p}=-i\hbar\frac{d}{dx}. $$ This is not a general expression for the momentum operator. It is the momentum operator written in a particular representation, the position ...


0

Yes, the terminology is sometimes a little bit sloppy. The Hilbert space is actually the product of the infinite dimensional Hilbert space defined on $\mathbb R^3 $ and two dimensional spinor space (or in relativistic qm, 4-dimensional space of Dirac spinors). My recommendation is to ignore terminology, and focus on mathematical structure. The wave function ...


5

I've learned that quantum wave functions can be described as a "ket vector" in an abstract vector space called Hilbert space. The position wave function, for example, used to express the probability of finding the particle at a point, can be described as a vector in an infinite dimensional Hilbert space. It seems you are talking about the position ...


0

If you quantize the manifestly covariant Polyakov action for the point particle, written in its 1st-order form as $$ S[x^{\mu}(\tau), p_{\mu}(\tau), \lambda(\tau)] = \int d\tau \left( p_{\mu} \dot{x}^{\mu} - \lambda \left( g^{\mu \nu} p_{\mu} p_{\nu} - m^2 \right) \right), $$ you will end up with wavefunctions which are functions over the spacetime like you ...


1

Its called a wave function because it comes in the form $\frac{\partial^2 u}{\partial t^2} = k \frac{\partial^2 u}{\partial x^2}$ (or rather, the multidimensional equivalent of that). That kind of equation models all sorts of waves, the way you and I think of waves. The randomness comes when one wishes to make an observation of the particle's state. The ...


1

The proposition "The wavefunction is all there is" is meant to clarify the difference between the many-worlds theory of quantum mechanics and some other quantum mechanical interpretations. In many other interpretations there are additional physical laws required to explain what happens during a measurement. In those interpretations there is, in ...


3

You’re exactly right: $|\phi(p)|^2$ gives the probability of measuring momentum $p$ at time $t=0$. An analogous relation holds for the time-dependent case: $$\Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dx e^{-i px/\hbar}\Psi(x,t)$$ This is simply due to the fact that one independently transforms between position and momentum space and between ...


2

This answer is simply to complement the one provided by J. Murray and discuss density functional theory, a much-used theory to study quantum systems with a 3-dimensional function: the electron density. For all N-electron systems, the kinetic energy and electron-electron interaction terms are the same. Therefore, all you need to specify such a system is the ...


3

No, there isn't really a loophole. Consider two particles which live in 1D, and let them be bosonic. The wavefunction of the composite system can most generally be written as some symmetric function $\psi(x_1,x_2)=\psi(x_2,x_1)$. If we make the assumption that $$\psi(x_1,x_2) = \frac{\psi_A(x_1)\psi_B(x_2) + \psi_B(x_1)\psi_A(x_2)}{\sqrt{2}}$$ and further ...


1

The Wolfram Functions site has the following integral which has the form of these matrix elements: $$\begin{align} \int_0^\infty &t^{\alpha-1}e^{-pt}L_m^\lambda(pt)L_n^\beta(pt)\,dt=\\ &\frac{p^{-\alpha}\Gamma(\alpha)\Gamma(n-\alpha+\beta+1)\Gamma(m+\lambda+1)}{m!\,n!\,\Gamma(1-\alpha+\beta)\Gamma(\lambda+1)}{}_3F_2(-m,\alpha,\alpha-\beta;-n+\alpha-\...


-1

The (complex) spherical harmonics is the solution of wavefunctions for the time-independent Schrödinger equation for the Coloumb potential, which is why we do care about it. It is through Bohr's footnote that we take the probability density as a function of the spherical harmonics, namely $|\psi |^2$, to obtain the orbitals of atoms you mention. Edit 1: As ...


0

I think there are two issues here: the wave function in a helix does not have to be periodic the wave function does not have to be real Periodicity After one tour of the helix we do not return to the same point, so the periodicity does not apply: $$\alpha(\phi+2\pi) = \alpha(\phi) + a2\pi\hat{k}.$$ What might be the source of confusion here is ...


0

Since, the case here is solved using Schrodinger's equation (a second order differential), we arrive at that particular solution theoretically as follows. In the case of Particle in a box, $\mathbf V(x) = 0$, for a particle inside the box. And hence, the Schrodinger equation becomes, $$\frac{d^{2}\psi}{dx^2} + \mathbf {k^2}\psi (x) = 0$$ where, $$\mathbf k =...


1

The eigenvalue equation obeyed by a particle in a box is a second order homogeneous linear differential equation of the form: $$ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}-E\psi(x)=0. $$ The answer to your question is purely mathematical. More generally, for a homogeneous linear differential equation of any order: $$ a_n\frac{d^ny}{dx^n}+a_{n-1}\frac{d^{n-1}...


3

Does anyone know how this function is arrived at? Was it experimental? Note that the wave function is not an observable and so cannot be arrived at experimentally. The Schrödinger equation for the 1D potential well is given by: $$-\frac{\hbar}{2m}\frac{\text{d}^2 \psi}{\text{d}x^2}=E\psi$$ Slightly re-written: $$\psi''+k^2\psi=0\tag{1}$$ where: $$k^2=\frac{...


Top 50 recent answers are included