Is energy conserved in quantum mechanics?

Question

if yes then how is it explained in case of the particle in a box problem ?

Additional Details

if yes then how is it explained in case of the particle in a box problem or in the case of tunneling through a potential barrier?

if yes then how is it explained in case of the tunneling through a potential barrier?

Answer 1

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Energy *is* conserved in quantum mechanics, in two ways:

(1) On average. Operators have no average rate of change if they commute with a special operator called the Hamiltonian, which is basically the "Total energy operator." Since all operators commute with themselves, the Hamiltonian commutes with itself, and therefore there is no average gain or loss of the Hamiltonian.

(2) In the long run. You can derive that, in non-average cases, sometimes a particle will seem to violate conservation of energy -- it may, for example, "tunnel" through a potential barrier, even if that potential barrier requires much more energy than the particle itself has. How does it get to the other side, if not by going through the middle? And how does it go through the middle if it doesn't have the energy?

The answer is that there is a *Heisenberg uncertainty* in the amount of energy a particle has, just as there's a Heisenberg uncertainty in where it is or what momentum it has. As momentum relates to space, so energy relates to time: and instead of dp dx being greater than hbar, you have dE dt being greater than hbar. In other words, the particle can briefly "steal" an energy dE from the vacuum, but it has to "give it back" after a time hbar/dE.

This is more thoroughly covered in Griffiths' Introduction to Quantum Mechanics.

There is no relevance of the Particle in a Box solution to this particular problem, as far as I know. With Particle-in-a-Box, you get an exact description of the energy-eigenfunctions with specific, constant energies -- no energy conservation problems required or obtained.

Answer 2

Energy is conserved in everything. Sorry but I have no idea what the box thing is.

Answer 3

Is Energy Conserved in General Relativity?

In special cases, yes. In general -- it depends on what you mean by "energy", and what you mean by "conserved".

In flat spacetime (the backdrop for special relativity) you can phrase energy conservation in two ways: as a differential equation, or as an equation involving integrals (gory details below). The two formulations are mathematically equivalent. But when you try to generalize this to curved spacetimes (the arena for general relativity) this equivalence breaks down. The differential form extends with nary a hiccup; not so the integral form.

The differential form says, loosely speaking, that no energy is created in any infinitesimal piece of spacetime. The integral form says the same for a finite-sized piece. (This may remind you of the "divergence" and "flux" forms of Gauss's law in electrostatics, or the equation of continuity in fluid dynamics. Hold on to that thought!)

An infinitesimal piece of spacetime "looks flat", while the effects of curvature become evident in a finite piece. (The same holds for curved surfaces in space, of course). GR relates curvature to gravity. Now, even in Newtonian physics, you must include gravitational potential energy to get energy conservation. And GR introduces the new phenomenon of gravitational waves; perhaps these carry energy as well? Perhaps we need to include gravitational energy in some fashion, to arrive at a law of energy conservation for finite pieces of spacetime?

Casting about for a mathematical expression of these ideas, physicists came up with something called an energy pseudo-tensor. (In fact, several of 'em!) Now, GR takes pride in treating all coordinate systems equally. Mathematicians invented tensors precisely to meet this sort of demand -- if a tensor equation holds in one coordinate system, it holds in all. Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles. In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects. (See the FAQ entry on black holes for some examples.)

These pseudo-tensors have some rather strange properties. If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime. By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation. For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy.

One other complaint about the pseudo-tensors deserves mention. Einstein argued that all energy has mass, and all mass acts gravitationally. Does "gravitational energy" itself act as a source of gravity? Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". So one can argue that "gravitational energy" does NOT act as a source of gravity. On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). So one can argue that "gravitational energy" IS a source of gravity.

In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes.

Let's look at four examples before plunging deeper into the mathematics. Three examples involve redshift; the other, gravitational radiation.

Very fast objects emitting light
According to special relativity, you will see light coming from a receding object as redshifted. So if you, and someone moving with the source, both measure the light's energy, you'll get different answers. Note that this has nothing to do with energy conservation per se. Even in Newtonian physics, kinetic energy (mv^2/2) depends on the choice of reference frame. However, relativity serves up a new twist. In Newtonian physics, energy conservation and momentum conservation are two separate laws. Special relativity welds them into one law, the conservation of the energy-momentum 4-vector. To learn the whole scoop on 4-vectors, read a text on SR, for example Taylor and Wheeler (see refs.) For our purposes, it's enough to remark that 4-vectors are vectors in spacetime, which most people privately picture just like ordinary vectors (unless they have very active imaginations).

Very massive objects emitting light
Light from the Sun appears redshifted to an Earth bound astronomer. In quasi-Newtonian terms, we might say that light loses kinetic energy as i

Source(s):

http://math.ucr.edu/home/baez/physics/Re…

Answer 4

In David Griffiths' Introduction to Quantum Mechanics he makes the point pretty strongly that energy conservation still is valid.

I remember also that somewhere in the book (maybe it's in one of the problems or footnotes) he says that the uncertainty principle gets misused and abused quite a bit, so when a physicist uses the uncertainty principle sometimes "it's best to keep a hand on one's wallet."

Answer 5

If you have a particle in an infinite well (or "box"), there is no tunneling. If the well is of a finite "height" (i.e. it takes a certain amount of energy to overcome it) then you will have a possibility of finding the particle a little bit outside the well. It never escapes, it just might be beyond the wall by a distance Dx more or less. If the particle has energy E and the box is of "height" Eb, the the extra energy the particle will need to match E is DE=Eb-E. If you solve the wave equation, you can find the average speed of the particle if its outside the box and from that you can estimate the time Dt that it spends outside the box. You will find that DE Dt will be of the order hbar/2. In other words, if you think of the particle being outside the box as a violation of the conservation of energy, it is clear that it can only last for time Dt, more or less.

Source(s):

http://en.wikipedia.org/wiki/Finite_pote…