A counter-example to the quantum interest conjecture

According to the quantum interest conjecture any negative energy pulse must be associated with a positive energy pulse of greater magnitude than that of the negative energy pulse. In this paper we will demonstrate a counter-example to this conjecture. We will show that, for a massless scalar field in 1-1D space-time, it is possible to generate an"isolated"negative energy pulse.


Introduction.
According to the quantum interest conjecture any negative energy pulse must be associated with a positive energy pulse of greater magnitude than that of the negative energy pulse. That is, an "isolated" pulse of negative energy cannot exist. The quantum interest conjecture was originally proposed by L.H. Ford and T.A. Roman [1]. This idea has been further developed and commented on by a number of others [2][3][4][5]. The quantum interest conjecture is closely associated with the concept of quantum inequalities [6,7].
The quantum inequalities are lower bounds on the weighted average of the energy density. It has been shown that the quantum interest conjecture can be derived from the quantum inequalities [3]. Counter-examples to the quantum inequalities have been claimed in two previous papers by the author [8,9].
In this paper we will demonstrate a counter-example to the quantum interest conjecture. In the following discussion a massless scalar field in 1-1 dimensional spacetime in the presence of a time varying delta function potential will be examined. The field operator   ,, , , V x t is the scalar potential. For this discussion   , V x t is given by, is the Dirac delta function and   t  is non-negative and is specified by, During the interval 0 t T is assumed to continuously decrease from its initial value 0  at 0 t  to its final value which is zero at t T  . It will be shown that during this interval a negative energy pulse is produced. The pulse is isolated and is not associated with a positive energy pulse.

Mode Solutions.
In this section we will solve Eqs. (1.1) where   , V x t is given by (1.2). The solutions are given by, The â  and † a  are the annihilation and creation operators, respectively.
There are both odd and even solutions to this equation. The odd solutions are of the form These solutions do not interest us due to the fact that they are not affected by the delta function scalar potential. This is because they are equal to zero at 0 x  . Therefore they do not contribute to the change in the energy and will not be considered further.
Solutions to (2.2) are considered in Ref [10]. It is shown in Appendix 1 that, using the results of Ref [10], the even solutions are given by, where A  is a normalization constant and the function Initially, when 0 is a constant and will be written as 0, B  . It is evident from (2.4) that, Therefore the initial solution, for 0 t  , can be written as, In addition to this assume the boundary conditions Use this in (2.8) along with some trig identities and the fact that The normalization constant A  is given by solving the normalization condition, From this we obtain,

Kinetic energy density.
The kinetic energy density operator is given by, The kinetic energy density operator is the same as the energy density operator at any point in space where the scalar potential is zero. This is due to the fact that the energy density operator includes a term containing the scalar potential. For the problem we are considering the scalar potential is zero everywhere except at 0 x  . This means that the kinetic energy density and energy density operators are identical except at this point.
When t T  the kinetic energy density and energy density are identical everywhere because, from (1.3), the scalar potential is zero.
In order to calculate the kinetic energy density expectation value, the state vector on which the field operators act must be specified. This state vector will be designated by 0 and is defined by the relationship ˆ0 0 a   .
From the above discussion the kinetic energy density expectation value is, This can be shown to be equal to, and where   is the kinetic energy density of a given mode.
There is a problem with evaluating (3.3) which is due to the fact it will be infinite.
However we are not really interested in the total energy density but only the change in the energy density with respect to the unperturbed vacuum state. To achieve this result we will use mode regularization. The kinetic energy density of each mode will be given by, being the kinetic energy density of the unperturbed state and is given by . The total regularized kinetic energy density is given by summing up the change in the kinetic energy density of each mode to obtain, This equation is derived in the Appendix 3. Note that is both space and time independent. Therefore we will drop the explicit dependence on x and t in the rest of this section.
Next, use (2.5) in the above to obtain, The "regularized" kinetic energy of each mode is, then, given by, In the limit that L   we can use This is evaluated in Appendix 2 to obtain, The total kinetic energy, K E , is the kinetic energy density, 00,R T , integrated over all space From the above results we see that the kinetic energy density, 00,R T , is effectively zero everywhere in the limit L   . However the total kinetic energy, K E , is positive.

Generating a negative energy pulse.
At this point we have established the initial state which is valid for 0 t  . Between 0 t  and t T  the scalar potential decreases to zero. In the following discussion we will show that this reduction in the scalar potential results in the launching of a pulse of negative kinetic energy density. To see why this makes sense consider the expression for the total kinetic energy   t   associated with a given mode. This is just the kinetic energy density of the mode integrated over all space. Referring to (3.4) and (3.5) we obtain, is implied. Take the time derivative of this expression and use (2.2) to obtain, This yields, where we have used the boundary condition The total change in the kinetic energy of the mode during this time interval is given by integrating the above quantity to obtain, What might this energy be? Assume that   Use this in (4.4) to obtain, Due to the fact that   t  is decreasing the quantity   d t dt  will be negative which means that will also be negative. The change in the total kinetic energy is summation of the change in the kinetic energy of each mode. This yields, This is negative because each What happens to the energy density during this time interval? Note that at t T  the energy density in the region outside of x T  cannot have changed because the effect due to the changing of the scalar potential cannot travel faster than the speed of light which we have taken to be equal to 1. Therefore all the energy generated during the interval 0 T t   is contained within the region x T  at the time t T  . This means that the average energy density within the region x T  is negative.
Next what happens for the time t T  for which   0 t   ? Referring to (4.3) the total kinetic energy density can no longer change. Therefore the negative energy that was produced during the time interval 0 T t   will move out at the speed of light and not be followed by a pulse of positive energy since, for t T  , no further energy (negative or positive) is being generated. The result is an isolated region of negative energy moving away from the origin at the speed of light. One half of the negative energy moves in the positive direction and other half moves in the negative direction.
These results are based on the validity of the quasi-static approximation. They will be verified in the next two sections when we consider exact solutions.

An exact solution I.
In this section we will show that a negative energy pulse is radiated for the case where is given by, Use this in (2.3) to obtain, Use this result and (5.1) in (4.4) to show that the change in kinetic energy during the interval 0 t  to t T  for a given mode is given by, For 0 t  the mode solutions are given by (2.6). For this case given by (2.5). Therefore for 0 t  , Rearrange terms to obtain, Therefore during the interval from 0 t  to t T  the change in the kinetic energy of each mode is negative. For t T  when   0 t   there is no additional change in the kinetic energy. The result is a pulse of negative energy moving out at the speed of light. This can be seen by direct examination of the kinetic energy density.

Using (2.3) and (2.4) in (3.4) the kinetic energy density of each mode, for
The details of this calculation are given in Appendix 3. The regularized energy density of each mode is, Use (5.13) to obtain, Using this result the total kinetic energy density is, We can drop all   2 1 O L terms from the integrand in the above expression. This allows . That is, we can replace  by 0  . Thus we obtain, where we have replaced the dummy variable 0  with  to simplify notation. Note that for this case, in the limit L   , 2 2 A L   in (5.16).

An exact solution II.
In this section we will solve for the energy density for the case where the scalar potential decreases continuously to zero. Let   t  be given by, Note that   Making the substitution u T x   we obtain, Refer to (5.16) and substitute 2 We use this result in (6.8) which is evaluated using numerical integration as discussed in is negative which is consistent with the analysis of Sections 4 and 5. T can be found by using the above values in (6.2).
The above result has been obtained by setting 0 x  in (5.13). However the same result would is obtained for 0 x  . Therefore at this point we have found that at t T  the total energy in the region T x  is negative. The average energy density in the region and is negative. In the region x T  the energy density is effectively zero.
When t T  the quantity   0 t   as specified in Eq. (6.1). The result of this is that no additional energy is produced. The kinetic energy density for the region t x T   is zero for the half space where 0 x  . This can be seen by referring to (5.16) which shows that the kinetic energy density of a given mode is zero when   0 t x    is zero which will be the case when t x T   . There is an analogous result for the half space