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 =...


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 ...


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 ...


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 ...


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, ...


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 $\...


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 ...


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

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 ...


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 ...


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 ...


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{...


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 ...


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

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 ...


2

Technically, a spherical coordinate system is defined in 3-space $\mathbb{R}^3\backslash\{0\}$ except the origin $r=0$. Therefore, spherical coordinates are a poor description of the system at the origin $r=0$. The free particle itself has no beef with the origin. There is no actual physical boundary conditions at the origin. The moral is that we shouldn't ...


2

The solutions for the radial part are in terms of spherical Bessel functions: \begin{align} R_\ell(\rho)=j_\ell(\rho) \end{align} This is done in details in Landau&Lifshitz's QM text. Your radial equation for $u$ should come out to be \begin{align} u''-\frac{\ell(\ell+1)}{r^2}+k^2 u(r)=0\, . \end{align} There's a boundary condition at $r=0$ that should ...


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 ...


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.


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 ...


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\...


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\...


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 ...


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 ...


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

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 ...


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/...


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 ...


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 ...


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