New answers tagged

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It can be easily seen using the Heisenberg picture. Take a Hermitian operator $A$ that commutes with the Hamiltonian $H$. Remember that the eigenvalues of $A$ are observables. Then $A$ also commutes with the time evolution operator $U(t) = e^{-i H t}$. $[A,U(t)] = 0 \quad \rightarrow AU(t) = U(t)A \rightarrow A = U(t)AU(t)^\dagger $ So the operator $A$ ...


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If you have an eigenstate of an operator $A$: $$A|Ψ(t=0)>=a|Ψ(t=0)>$$ Then apply $A$ to the time evolution of the wavefunction: $$A|Ψ(t)>=Ae^{-iHt}|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=a|Ψ(t)>$$ Note that the second to last equality is only true if $[A, H]=0$. So the eigenstate of $A$ remains an eigenstate of $A$! This applies ...


2

A conserved quantity is one that commutes with the Hamiltonian for the simple reason that $[A,H] = 0$ implies $$ \frac{\mathrm{d}}{\mathrm{d}t} A = 0$$ in the Heisenberg picture. Another way to see that commuting with the Hamiltonian means conservation is to consider that the time evolution operator $U(t) = \exp(-\mathrm{i}Ht)$ is just the exponential of ...


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Note that pendulum motion is only sinusoidal for small angular displacements; as you increase the amount of swing the harmonic approximation fails. Lagrangian mechanics gives you a handle on all of the cases.


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The point is that your equation (1) does not make any sense. On the left side, you have operator acting on a vector in some abstract vector space. On the right side you have the position representation of momentum operator acting again on the same abstract vector. Those objects live in different spaces, you cannot just multiply them (actually you can if you ...


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may be you can think this way. The wavefunction $|x\rangle=e^{i(kx-\omega t)}$ How do we extract the momentum $\hbar k$ out of it by some operator ? $$ \frac{\partial}{\partial x}|x\rangle=ik|x\rangle $$ to get $\hbar k$, you need to either multiply the operator with $-i\hbar$ or with $\frac{\hbar}{i}$. So the momentum operator is $$ ...


4

OP's ket first equation$^1$ $$\tag{1} \hat{p}|x\rangle ~=~+i\hbar\frac{\partial |x\rangle}{\partial x}$$ is explained in eq. (7) of my Phys.SE answer here. In short, eq. (1) is consistent with the corresponding bra equation $$\langle x |\hat{p} ~=~-i\hbar \frac{\partial \langle x |}{\partial x} $$ via Hermitian conjugation. The bra equation in turn is ...


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To hit a rigid body such that it rotates about a specified point, you need to hit it at the instant center of percussion. If the pivot point is a distance $c$ from the center of mass, then the percussion center is located a distance $$\ell = \frac{I_{cm}}{m c} = \frac{\kappa^2}{c} $$ away from the center of mass. Here, $I_{cm}$ is the mass moment of inertia ...


1

If a ball of say radius $R$ rolls without slipping it has both linear ($p$) and rotational momentum ($L$): $$p=mv$$ $$L=I\omega$$ Where $m$ is mass of ball, $I$ is inertial moment of ball, $v$ is translational (linear) speed and $\omega$ is angular speed. For rolling without slipping the following condition also holds: $$v=\omega R$$ The ball will have ...


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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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The small object exerts a force in the opposite direction to the normal force on the cart


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It depends on if the object being placed is moving at the same exact speed and direction as the car when placed inside. If so then nothing will happen but if different then the speed or direction of the car will change.


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If we observe the path of the ball as radius reduces, we can see that the path followed is not circular. So between the radius and tangent, angle formed will not be 90 degrees.. This creates a component of tension along the path of the ball thus increasing its velocity


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Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe. Conservation means that these quantities cannot spontaneously change. Let's ...


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In a collision momentum is conserved if there are no external forces. enefgy is also cionserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones when kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in ...


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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 ...


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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 ...


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"velocity would stay constant, mass would decrease and so the momentum of the trolley decreases" , is a correct argument , but if this is a system where gravitational forces are present , you have neglected the pressure on water flowing through the hole in the trolley .The pressure on water flowing through the hole will produce a velocity(1) in it ...


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This is all to do with the centre of mass motion and the fact that there are no external forces for the hemisphere & particle system. The velocities are measured relative to the ground and as you started with no momentum in the x-direction (other than that due to the infinitesimal impulse) you must end up with no net momentum in that direction. Later ...


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In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $ V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$


1

Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...


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This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...


1

I would say the answer is rather simple. If you do various experiments to measure the position (or momentum) of a particle, then QM only gives you consistent and accurate predictions when the mathematical symbols $x$ and $p$ are associated with the measured position or momentum. However, position and momentum are not the only canonically conjugate operators ...


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Momentum and inertia are closely related properties. Newton's first law states that an object will continue in a straight line with constant momentum if no net external forces act on it. When you apply the brakes, the road applies a net external force on the car-plus-wheels. This force will cause the car to slow down, and the road to speed up (conservation ...


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The answer may be deeper than you expect. The operator $K$ that the OP defined --- here it will be called $-i\nabla_x$ --- is the generator of space translations. And the operator $X$ --- here called $x$ --- is the generator of momentum translations. If you exponentiate them, to form the corresponding one-parameter unitary groups of translation, you get ...


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It is not about momentum. The total momentum is conserved: the fingernail flies in one direction, and you (plus the nail clipper) suffer a recoil to the opposite direction. The thing is that your piece of fingernail has a very small mass compared to your body mass, and the firing speed is actually not so high, so the recoil is negligible. About why the ...


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This is essentially the same question as https://astronomy.stackexchange.com/questions/13302/ In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum is $\text{m1} \text{v1}$ and the initial kinetic energy is $\frac{\text{m1} \text{v1}^2}{2}$, since m2 is at rest. Let u1 and u2 be the velocities of the ...


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A test particle in general relativity moves along a worldline. If the world line is differentiable, then it has a tangent. If the test particle is similar to a massive particle, that worldline has tangents that are always timelike. And you could make a unit tangent. That unit tangent lives in the 4d tangent space. As does the energy-momentum vector. In ...


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Let us consider the cart as one ball and the sand as another ball moving together. The answer could easily be derived using conservation of momentum but since you have doubts, I will begin with this explanation. When the two ball are separated, the sand ball does not change its speed but continues with the same speed. Imagine this, two balls moving closely ...


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$F = ma$ and the only forces on the cart are gravity and the reaction force of the surface that balance. So the cart does not accelerate: it keeps moving at constant velocity.


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It relies on conservation of energy and momentum and the equation for energy in special relativity: $E^2 = (pc)^2 + (mc^2)^2$. Here you go. Energy of photon: $E_\gamma = \hbar\omega = p_\gamma c$, where $p_\gamma$ is the momentum of the photon. Assume the electron is initially at rest, so it's energy is simply $m_ec^2$. By conservation of energy, the ...


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As another hint to your problem, you'll also want to consider conservation of energy (there's a big clue in your problem that conservation of energy is important --there is no friction between the cart and the path). To start you off, here are the important conservation of energy equations for your problem: $$E_k = \frac{1}{2}mv^2= ...


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Conservation of momentum is a little inconvenient under gravity. Anyway, the last sentence in the question means that $p = mv$, throwing each sand bag gives the same momentum boost. Let's name it $\Delta p$. (you might need to work that out since your teacher might ask for a detailed explanation; I'll leave that to you) Now your question asks, how to ...


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In addition to the arguments already given (units), the 4-momentum $p^\mu$ is defined from the 4-velocity $u^\mu$ via: $$ p^\mu = m u^\mu $$ with $u^\mu = \frac{d x^\mu}{d\tau} = \gamma \frac{dx^\mu}{dt} =\gamma (c, \vec{v})$. Hence: $$ p^\mu = (\gamma mc, \gamma m\vec{v}) = (E/c, \vec{p})$$


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The uncertainty principle states that product of the standard deviations of two observables is bounded below by a multiple of the absolute value of the expectation value of the commutator of the two operators. In detail, if $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$ and $\Delta B=\sqrt{\langle ...


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The velocity $\hat v$ is a well-defined operator for particles, equal to $\hat p/m$ in non-relativistic physics. The non-relativistic, non-linear correction factors could be added but that would create a can of worms because the correct way to describe relativistic particles requires quantum field theory where $\hat x,\hat v,\hat p$ aren't quite well-defined ...


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They are consistent with each other because quantum mechanical momentum is not the change in position. There is no quantum notion of velocity. Classically, velocity is the time derivative of $x(t)$ along a particular trajectory. The quantum theory has no notion of a real-valued trajectory $x(t)$. Quantum states, in general, are not position eigenstates, ...


1

Two things are required to prove this. First, you haven't really defined your state $|p\rangle$ in its entirety since you haven't defined what normalization you are using for this state. In the formula above, it seems to me that you are using the normalization $$ \langle {\bf p} | {\bf p}' \rangle = \frac{2 E_p }{ 2\pi } \delta^3 ( {\bf p} - {\bf p'} ) $$ ...


2

Your first part was correct; but for the second part, you have to equate the energy of A plus the stored energy in the spring to the energy of B (because you start with no energy in the spring, and all the energy as kinetic energy in B). So the expression for the stored energy is $E_\mathrm{spring}=\frac12 k x^2 = \frac12 m_b v_b^2 - \frac12 m_a v_a^2$. ...


3

Well I would say your first approach is absolutely correct and will result in right answer if the calculations are done correctly (I have not done the calculations) but your second approach is faulty. Problem with second approach In the second approach you have used the conservation of mechanical energy it seems. The problem is that conserving mechanical ...



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