# Does the uncertainty principle violate the law of conservation of energy?

What is the scientific view of the beginning of universe? Quantum fluctuation seems to contradict with the law of conservation of energy.

Uncertainty Principle does seem to violate the Law of conservation of energy, I read it on this Wikipedia page and got confused.

• can you give a link of this? – anna v Feb 25 '14 at 14:12
• Please edit my question,I am very poor at English. – Tom Lynd Feb 25 '14 at 15:26
• @ anna: I have provided a link. – Tom Lynd Feb 25 '14 at 15:28
• Possible duplicate: physics.stackexchange.com/q/55860/2451 – Qmechanic Feb 25 '14 at 20:57
• The time-evolution of an operator $\hat{O}$ in Heisenberg picture is given by $$\frac{d \hat{O}}{dt} = \frac{ [\hat{O},\hat{H}]}{i \hbar}$$. If $\hat{O}=\hat{H}$, $[\hat{H},\hat{H}]=0$ the energy should be conserved. I guess for $\Delta E \Delta t \approx \hbar$. The fluctation comes from interaction between matter and radiation field, and the total energy (matter + radiation field) should be conserved. – user26143 Feb 25 '14 at 22:12

This is an experimental physicist's answer:

The linked article is careful to state:

That means that conservation of energy can appear to be violated, but only for small values of $t$ (time)

Italics mine.

Conservation of energy is an experimental fact that has been validated in innumerable experiments. This means, as far as the correspondence with quantum mechanical formulations goes, the energy is an observable and to every observable there corresponds a quantum mechanical operator, which, operating on the wave function of the system under observation, gives the value of the observable. Quantum mechanical operators either commute, anticommute or are non commuting and when they are not commuting a corresponding Heisenberg Uncertainty relation will be defined for the expectation values from the non-commuting operators. The Heisenberh Uncertainty Principle is a generic expression of the non-commutability

This allows the creation of particle-antiparticle pairs of virtual particles.

virtual means "as if", and it is a mathematical construct which affects the calculations for cross sections and lifetimes and other measurable quantities but cannot be measured by itself. In these simple Feynman diagrams: There are real particles and virtual particles, Real are the incoming and outgoing lines from the vertices, and virtual are the internal lines connecting the vertices. The real lines represent real particles on their mass shell. The virtual lines carry the quantum numbers of the particles with their name but are not on the mass shell, i.e. you cannot apply momentum and energy conservation at the same time because the mass corresponding to the virtual particle is arbitrary (well depends on probabilistic mathematical functions but arbitrary for the argument). Look at the diagram responsible for weak decays: the mass of the $W$ is $80.4\ \mathrm{GeV/c^2}$ and is the virtual particle exchanged in the beta decay of the neutron whose mass is less than $1\ \mathrm{Gev/c^2}$.

So when one reads "virtual particle creation" coupled with "apparent energy non conservation" this is what it means, that one cannot apply the conservation sums in the internal lines that hold when balancing input and output lines. They are a mathematical convenience to allow calculation of complicated integrals which the Feynman diagrams describe.

The effects of these particles are measurable, for example, in the effective charge of the electron, different from its "naked" charge.

This is a tautology, because virtual particles in the Feynman diagrams are necessary to make the calculations fit the observations, but one should keep in mind that they are a mathematical convenient representation that cannot be directly measured and thus is not a real situation. One can only measure input and output in an experiment. The calculations fit the experiments well and some people tend to treat virtual particles as something real, which leads to unnecessary confusion on conservation laws.

I do not agree with this answer. The energy and momentum conservation still holds at each vertex, as usual. the thing is that with the internal lines, the energy and momentum of the particle can take any value. For example take the exchanged pion and let's give it a momentum $k$. If the incoming proton has a momentum $p$, the outgoing will have momentum $q=(p-k)$. On the other vertex the incoming neutron has moment $p_1$, then the outgoing will have $p_2=p_1+k$. You see now that $k$ can take any value and there cannot be direct measurement on it, so it can be bigger that the rest mass of the proton (as in the $W$ case).