# How long can a virtual particles exist in a vacuum state?

As much as I have understood, virtual particles keep appearing and disappearing in a vacuum state. This leads to zero-point energy, Casimir effect etc. Now, I want to know if there is a lifetime (i.e. a time limit) for which these virtual particles can exist before disappearing. Also can there be a mechanism by which they can be made to stay longer?

• I guess you could think about the off-shell width. Nov 8 '16 at 12:32
• I guess, you can estimate it with Heisenberg's uncertainty relation. If your virtual particle has an energy $\Delta E$, it will exist for $\Delta t =\hbar/\Delta E$. However, it is possible that others might tell that virtual particles have only a formal sense, i.e. cannot exist in a proper sense. Nov 8 '16 at 13:49
• please look at this answer of mine and links therein physics.stackexchange.com/questions/286721/… Nov 8 '16 at 16:23

The concept of virtual particles is simply an analogy which resembles the underlying phenomena and which is often used to explain the QFT effects such as Casimir or Hawking radiation to beginners.

Reality is a bit different.

### Classical vacuum

In classical field theories we have a well-defined notion of "vacuum": a background spacetime with no field content. For instance, for a scalar field $\phi(t, \vec{x})$:

$$\phi(x^{\mu}) = \phi(t, \vec{x}) = 0$$

defines the classical vacuum configuration. Equivalently, it can be defined as a point on the phase space:

$$\phi(\vec{x}) = 0, \quad \pi(\vec{x}) = 0,$$

where $\pi(\vec{x})$ is the conjugate momenta density field (canonically conjugate to $\phi(\vec{x})$). Note how the two definitions of the vacuum are equivalent: you can take the phase space point defined by $\phi(\vec{x}) = \pi(\vec{x}) = 0$ and evolve it using Hamilton's equations to get $\phi(t, \vec{x}) = 0$.

### Uncertainty relation

In quantum mechanics, however, everything is much more complicated. Remember the renowned uncertainty relation? For any two canonically conjugate variables $x$ and $p$ we have $$\Delta x \cdot \Delta p \sim \hbar.$$

Lets apply this to the theory of a scalar field. Suppose that we know with absolute certainty that $$\phi(\vec{x}) = 0.$$

But this means that we cannot be sure of the value of $\pi(\vec{x})$! Moreover, Hamilton's equations will use this undefined $\pi(\vec{x})$ to take the field variable $\phi(\vec{x})$ to an undefined value in an arbitrary short interval of time.

This observation lies in the heart of the QFT treatment of vacuum: the uncertainty relation does not permit us to have a classical vacuum state, where the value of the field is zero at all spacetime points.

### QFT vacuum

Now to the mathematics of QFT. For free field theories we are able to define the QFT vacuum state, which we usually denote by $\left| 0 \right>$. As follows from the previous section, it does not have the properties of the classical vacuum state, especially, we can measure nonzero values of $\phi$ in this state!

The actual wavefunction for $\left| 0 \right>$ is the multi-dimensional generalization of the Gauss wavepacket:

$$\left| 0 \right> [\phi(\vec{x})] = \exp \left[ i \cdot \int d^3 x \, \alpha \, \phi(\vec{x})^2 \right],$$

where $\alpha$ is a dimension-full parameter which depends on the mass of the field and on $\hbar$.

It can be seen from this integral that the expectation (average value) of the field variable observable is $$\left< \phi(\vec{x}) \right> = \left< 0 \right| \phi(\vec{x}) \left| 0 \right> = 0,$$

which is what we expect from the vacuum state. However, the expectation of the field observable squared is nonzero:

$$\left< \phi(\vec{x})^2 \right> = \left< 0 \right| \phi(\vec{x})^2 \left| 0 \right> \neq 0.$$

This is proportional to $\hbar$ and thus extremely small in the classical approximation (I am ignoring the divergence issues here for simplicity). That is why we can say that in the classical limit $\left| 0 \right>$ is observationally equivalent to the classical vacuum.

$$\Delta t \cdot \Delta E \sim \hbar$$
is useful to estimate the average time scale relevant to the effects connected to virtual particles with energy scale $\Delta E$. However, this is as far as it goes: a useful tool for estimating the average time scale. It is wrong to interpret $\Delta t$ as a lifetime of the virtual particle.
• The last part about the interpretation of $\Delta t$ is important! This question is our canonical discussion of what the $\Delta t$ in the "time-energy uncertainty relation" actually denotes. Nov 8 '16 at 16:25