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If you want to use the theory of probability, a necessary condition for a wavefunction to be physically meanigful is $$\psi \in L^2(\mathbb{R}^3,d^3x)\:.$$ That is because, as a basic postulate of QM, we have that:
$\qquad\qquad\qquad\qquad$ $|\psi(x)|^2$ is the probability density to find the particle at $x$,
and the total probability must be $1$:
$$\int_{\...
7
There are different equivalent formulations of quantum mechanics, and how an interaction is represented in each formulation is different.
In the Schrodinger picture, operators are time independent and the state (or wavefunction) $\Psi$ depends on time. The Schrodinger equation governs the dynamics of the wavefunction:
\begin{equation}
i \hbar \frac{\partial ...
6
Yes, it's perfectly possible.
Start off with the obvious Ansatz,
$$
\psi(x)
=
\begin{cases}
A e^{\kappa x} & x<-a \\
B e^{\kappa x} + Ce^{-\kappa x} & -a<x<a \\
D e^{-\kappa x} & a<x
\end{cases}
$$
and then set up the boundary conditions,
\begin{align}
Ae^{-\kappa a} & = Be^{-\kappa a}+Ce^{\kappa a} \\
Be^{\kappa a}+Ce^{-\kappa a} ...
6
This is where the textbooks, in a way, lie to you. The operator $\hat{x}$ (and its counterpart, $\hat{p}$) is not a "good" quantum operator for a number of reasons, including the fact that these operators do not have normalizable eigenvectors, as you have seen. In particular,
$$|x\rangle$$
is not a sensible eigenvector as it is not normalizable. ...
5
The problem with my question was I assumed that this below postulate of quantum mechanics (given in R. Shankar chapter 4 as 3rd postulate) to be literally true
If the particle is in a state $|\psi\rangle$, measurement of the variable (corresponding to) $\hat{\Omega}$ will yield one of the eigenvalues $\omega$ with probability
$P(\omega)\propto |\langle \...
5
I think what you just discovered is that every (separable*) infinite-dimensional Hilbert space $\mathcal H$ is isometrically isomorphic to $\ell^2$. The isomorphism is given by mapping a vector $|\psi\rangle \in \mathcal H$ to the coefficient list $( \langle e_n | \psi \rangle )_n$.
In particular, $L^2(\mathbb R^d)$ is isomorphic to $\ell^2$. For example, ...
4
But the first postulate of quantum mechanics also says that |ψ⟩ completely describes the isolated system.
$|\psi\rangle$ completely describes the state of an isolated system at that moment in time. It does not describe how that state evolves with time.
What then determines H of an isolated system?
One could ask the same about a classical system as well. ...
3
There is no a-priori sense in having an infinitely large potential, you can't calculate with infinite numbers (except for some very special mathematical considerations). So, the physical meaning that is given to such a potential must be implicitely contained in the accompanying text that explains the infinite potential well. An this meaning is commonly: &...
3
For the first part, you would use Bloch's Theorem.
For the second part, why don't you plug it into the Schrodinger Equation and see what happens?
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The Schrödinger Equation holds for all the cases (assuming no relativistic corrections) including the case of the free particle. Recall the equation reads:
$$i\hbar\frac{\partial}{\partial t} \Psi(x,t) = \left [ - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\right ] \Psi(x,t) \, $$
where we want to solve for $\Psi$. The differences of the ...
3
It is perhaps a question of semantics, to some extent, but you can't talk about the energy of a quantum particle unless it is in a state of definite energy i.e. an eigenstate of the Hamiltonian. If you had a superposition of two energy eigenstates, say $$|\psi\rangle = c_1 |E_1\rangle + c_2 |E_2\rangle,$$ any measurement of the energy will result in either ...
2
I don't know exactly what you mean with 'general'. For your specific example, we find that the eigenvalues of the Hamiltonian $E_n$ are given by the equation in your post. Hence, the corresponding eigenfunctions fulfill $\hat{H} \psi_n(x) = E_n\, \psi_n(x)$. As you pointed out, in the example a solution of the time-dependent Schrödinger equation is given by $...
2
Noether's theorem says that if the Lagrangian of your system is invariant under translations then momentum is conserved. Since the potential $U(x)$ is not translation invariant the Lagrangian isn't either. So we shouldn't expect momentum to be conserved. For a classical particle moving in a potential $U(x)$ this is true as well. The particle continuously ...
answered Feb 24 at 22:17
AccidentalTaylorExpansion
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2
There are two different concepts in here that I think you are mixing up. One is orthogonality, the other is linear independence. States that are orthogonal are necessarily linearly independent, but not all linearly independent states have zero overlap (i.e. are orthogonal).
The solution you posted does not say that $\psi_\text{A}$ and $\psi_\text{B}$ are ...
2
Let's write wave functions in these three regions for a level of energy $-E, E>0$. The corresponding exponent of wavefunction is $q = \sqrt{\frac{2 m E}{\hbar^2}}$:
Region 1: $x < -a$
$$
\Psi_1(x) = A \exp(qx).
$$
Region 2: $-a < x < a$
$$
\Psi_2(x) = B \exp(qx) + C \exp(-qx);
$$
Region 3: $x > a$
$$
\Psi_3(x) = D \exp(-qx);
$$
Then ...
2
The position eigenstates are not actually part of any Hilbert space at all. A position eigenstate wave function with eigenvalue $x_{0}$ is $\psi(x)=\delta(x-x_{0})$. If we try to normalize this, we need to calculate
$$\int_{-\infty}^{+\infty}dx\,|\psi(x)|^{2}=\int_{-\infty}^{+\infty}dx\,[\delta(x-x_{0})]^{2}=\delta(0)=\infty.$$
So this basis state is ...
2
This answer is meant as a supplement to Valter Moretti's one.
Quantum systems are defined by their Hamiltonian, however, this is normally not their single observable. We have the basic observables stemming from the space-time context (coordinate, momentum, spin), plus the Hamiltonian, plus other observables stemming from other symmetries (orbital / total ...
2
It is a mathematical phenomenon, it happens due to time-dependent Schroedinger equation. It is similar to diffusion equation, localized peaks in tend to dissipate according to both equations. So any localized wavepacket spreads unless that spreading is stopped by a potential rise.
This behaviour can be most easily shown and quantified for free particle. If ...
2
if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle?
No, when you measure the energy of a (not necessarily free) particle you find an interval of values due to the finite precision of every measurement apparatus.
If you know that ...
2
Note that the RHS of your equation is indeed zero: In general, we evaluate the time derivative of the expectation value of an operator $A$ in the state $\Psi$ as
$$ \frac{\mathrm{d}}{\mathrm{d}t} \langle A \rangle_{\Psi} = \frac{-i}{\hbar} \langle[A,H]\rangle_{\Psi} + \langle\partial_t A\rangle_{\Psi} \quad . $$
For a stationary state $\psi$, we have that $H\...
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The answer is almost contained within your question itself: if the left-hand side is zero, then the right-hand side must be zero as well. One way you can understand why the right-hand side is zero from a stationary state would be to use a WKB-esque intuition: if we're in a steady state, then the probability density of particles is not building up or ...
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$\require{cancel}$I am pretty sure you can demonstrate this with contradiction. You can however prove this directly as follows:
Let $\psi(x)$ be a real and normalized eigenfunction of the Hamiltonian $H$. It is a known theorem that the eigenvalue $E$ that corresponds to $\psi(x)$ is the expected value of $H$ on the eigenfunction itself. Thus,
$$E = \langle H\...
2
Wavefunctions in quantum mechanics are elements of the Hilbert space $L^2(\mathbb{R})$. This is usually described as the vector space of square-integrable functions with the inner product:
$$\langle \psi,\phi \rangle = \int_\mathbb{R} \psi^*(x)\phi(x) dx$$
However one should note that one of the axioms for an inner product is that the inner product of any ...
1
The equation $\psi'' = +A(x)\psi$ can be interpreted as the curvature of the wavefunction, $\psi''$, having the same sign as the wavefunction, $\psi$.
That means when the wavefunction is positive, $\psi > 0$, the function has positive curvature, $\psi'' > 0$, like a smile. But that implies as we go off towards positive and negative infinity in the $x$ ...
1
All the literature says that the physically meaningful solutions to
the Schrödinger equation in an infinite potential well must fulfill
the boundary condition that the wave function is 0 at the walls of the
well, otherwise the wave function wouldn't be continuous.
That's not the reason at all: there are plenty potentials $V(\mathbf{r})$ for particles with '...
1
Assuming by "spread" you mean how "wide" the distribution of $\mid \psi(x,t) \mid^2$ (or for momentum space, the spread of $\mid \psi(k,t) \mid^2$).
Quantum mechanics states that the wave packet of a particle "spreads-out" in position again after a measurement on this particle has been made.
Before and after measurement, the ...
1
The thing that the Hilbert-space vector $\vert \psi\rangle$ describes is, as the name suggests, the state of the system at a particular point in time. More precisely, $\vert \psi\rangle$ tells you about the statistics of any observable you'd want to measure at that point in time.
This has nothing to do with describing how the state of the system (i.e. $\vert ...
1
It is always good to have several ways to approach a problem, not least because it can give you new insights or allow you to solve other types of problems. Moreover, certainly imo, the path integral approach is conceptually much nicer and very logical. If you want a quick intro to path integral I suggest you read the first chapter of Zee’s QFT in a nutshell. ...
1
The solution you reference only works if you already know what the energies are and the energy is the only quantum number needed to describe the Hilbert space. In general, we won't know ahead of time what the possible energies are, what states they correspond to if we do, or anything what other quantum numbers are attached to the states of our Hilbert space (...
1
The problem is likely to be that you used eV as a scale for energy, but you also used kg as a unit of mass, which means you're mixing different units and there should be a conversion factor somewhere ($1$ eV $\approx 1.6 \times 10^{-19}$ J $= 1.6 \times 10^{-19} \, \mathrm{kg \cdot m^2 \cdot s^{-2}}$).
By instead using only SI units $V_0 = 250.6 \, \mathrm{...
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