Conjugates to one particle Hamiltonians in 1-dimension in differential form

Abstract

A time operator is a Hermitian operator that is canonically conjugate to a given Hamiltonian. We construct such operators in position representation for a 1-dimensional particle. The construction is first simplified by assuming a definite form for the kernel that is based on the free particle case and is justified by the correct classical limit of the operator. This leads to a family of Hamiltonian conjugates that can be derived by finding a twice-differentiable function using a hyperbolic second-order partial differential equation with appropriate boundary conditions. Additional conditions may be imposed to produce different Hamiltonian conjugates such as those corresponding to time of arrival operator. A larger solution space of Hamiltonian conjugates, like those that can arise from kernels involving Dirac Deltas, can be also constructed by removing the simplifying assumption and treating the operators as a distribution on some function space.

Appendix 1 The General Potential Solution

The general potential (93) for the time kernel equation (18) gives the recurrence relation (94) for the coefficients of the analytic solution. The boundary conditions (87) give two branches of nonvanishing terms: the first condition gives the time of arrival, and the second condition gives the shifting term. This second condition gives a nonvanishing $\alpha _{0,2N-1}$ for positive integer N, and so the recurrence relation (94) makes all $\alpha _{m,n}$ terms with even n vanish.

We solve this following analogous steps as that in [15]. We now look at the branch of coefficients $\alpha _{m,n}$ with odd n. First, we have, for $m \ge 0$,

$$\begin{aligned} \alpha _{m,2N-1} = -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \delta _{m,0} \, . \end{aligned}$$

(163)

Using the recurrence relation (94), we get $\alpha _{m,2N+1}$ for $m \ge 1$,

$$\begin{aligned} \alpha _{m,2N+1} = -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{\mu }{2\hbar ^2} \frac{1}{2N+1} \frac{a_m}{2^{m-1}} \, . \end{aligned}$$

(164)

Similarly,

$$\begin{aligned} \begin{aligned} \alpha _{m,2N+3}&= -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \left( \frac{\mu }{2\hbar ^2}\right) ^2 \frac{1}{m(2N+1)(2N+3)} \sum _{s=1}^{m-1} \frac{s a_s a_{m-s}}{2^{m-2}} \\&\qquad -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{\mu }{2\hbar ^2} \frac{1}{m(2N+3)} \frac{a_{m+2}}{2^{m+1}} \left( {\begin{array}{c}m+2\\ 3\end{array}}\right) \, . \end{aligned} \end{aligned}$$

(165)

This suggests that, for $m \ge 1$ and $j \ge 1$,

$$\begin{aligned} \alpha _{m,2N-1+2j} = \sum _{s=0}^{j-1} \left( \frac{\mu }{2\hbar ^2}\right) ^{j-s} \alpha _{m,j}^{(s)} \, . \end{aligned}$$

(166)

The recurrence relation (94) gives

$$\begin{aligned} \alpha _{m,2N-1+2j} = \frac{\mu }{2\hbar ^2} \frac{1}{m(2N-1+2j)} \sum _{s=1}^\infty \frac{a_s}{2^{s-1}} \sum _{k=0}^{[s]} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \alpha _{m-s+2k,2N+2j-2k-3} \, . \end{aligned}$$

(167)

Note that $\alpha _{m-s+2k,2N+2j-2k-3}$ is nonvanishing for $m-s+2k \ge 0$ and $2N+2j-2k-3 \ge 2N - 1$. Also note that the binomial coefficient is nonvanishing for $s \ge 2k+1$. We can then rewrite the above equation into

$$\begin{aligned} \alpha _{m,2N-1+2j} = \frac{\mu }{2\hbar ^2} \frac{1}{m(2N-1+2j)} \sum _{k=0}^{j-1} \sum _{s=2k+1}^{m+2k} \frac{a_s}{2^{s-1}} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \alpha _{m-s+2k,2N-1+2(j-k-1)} \, . \end{aligned}$$

(168)

Substituting (166) to the right-hand side gives

$$\begin{aligned} \begin{aligned} \alpha _{m,2N-1+2j}&= \frac{\mu }{2\hbar ^2} \frac{1}{m(2N-1+2j)} \sum _{k=0}^{j-1} \sum _{s=2k+1}^{m+2k} \frac{a_s}{2^{s-1}} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \\&\quad \times \sum _{r=0}^{j-k-2} \left( \frac{\mu }{2\hbar ^2}\right) ^{j-k-1-r} \alpha ^{(r)}_{m-s+2k,j-k-1} \, . \end{aligned} \end{aligned}$$

(169)

Since $0 \le k \le j - 1$, then the summation in r is nonvanishing up to $j-k-1$, and so we can rewrite this into

$$\begin{aligned} \begin{aligned} \alpha _{m,2N-1+2j}&= \frac{1}{m(2N-1+2j)} \\&\quad \times \sum _{k=0}^{j-1} \sum _{r=0}^{j-k-1} \sum _{s=2k+1}^{m+2k} \frac{a_s}{2^{s-1}} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \left( \frac{\mu }{2\hbar ^2}\right) ^{j-k-r} \alpha ^{(r)}_{m-s+2k,j-k-1} \, . \end{aligned} \end{aligned}$$

(170)

Interchanging the r and k summations, we get

$$\begin{aligned} \begin{aligned} \alpha _{m,2N-1+2j}&= \frac{1}{m(2N-1+2j)} \\&\qquad \times \sum _{r=0}^{j-1} \sum _{k=0}^{j-r-1} \sum _{s=2k+1}^{m+2k} \frac{a_s}{2^{s-1}} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \left( \frac{\mu }{2\hbar ^2}\right) ^{j-k-r} \alpha ^{(r)}_{m-s+2k,j-k-1} \, . \end{aligned} \end{aligned}$$

(171)

Rewriting this by letting k run from 0 to r gives

$$\begin{aligned} \begin{aligned} \alpha _{m,2N-1+2j}&= \sum _{r=0}^{j-1} \left( \frac{\mu }{2\hbar ^2}\right) ^{j-r} \\&\qquad \times \frac{1}{m(2N-1+2j)} \sum _{k=0}^{r} \sum _{s=2k+1}^{m+2k} \frac{a_s}{2^{s-1}} \left( {\begin{array}{c}s\\ 2k+1\end{array}}\right) \alpha ^{(r-k)}_{m-s+2k,j-k-1} \, . \end{aligned} \end{aligned}$$

(172)

Comparing this with (166), we thus get the recurrence relation

$$\begin{aligned} \alpha _{m,j}^{(s)} = \frac{1}{m(2N-1+2j)} \sum _{r=0}^{s} \sum _{n=2r+1}^{m+2r} \frac{a_n}{2^{n-1}} \left( {\begin{array}{c}n\\ 2r+1\end{array}}\right) \alpha ^{(s-r)}_{m-n+2r,j-r-1} \, . \end{aligned}$$

(173)

From (166), we see that the nonvanishing coefficients contribute to $v^{2N-1+2j}$, giving a Wigner-Weyl contribution (20) proportional to

$$\begin{aligned} {\mathcal {T}}_\hbar \propto \frac{1}{\hbar } \alpha _{m,2N-1+2j} \int _{-\infty }^\infty v^{2N-1+2j} {{\,\mathrm{sgn}\,}}(v) \exp \left( -i\frac{p}{\hbar }v\right) \hbox {d}v \, , \end{aligned}$$

(174)

which gives, upon using (24),

$$\begin{aligned} {\mathcal {T}}_\hbar \propto \sum _{s=0}^{j-1} (-1)^j \left( \frac{\mu }{2}\right) ^{j-s} (2N-1+2j)! \alpha ^{(s)}_{m,j} \frac{\hbar ^{2N-1+2s}}{p^{2j}} \, . \end{aligned}$$

(175)

From our results for linear systems, we infer that the Wigner-Weyl transform here should look like

$$\begin{aligned} {\mathcal {T}}_\hbar = \beta \frac{(2N-1)!}{2^{N-1}} \frac{\hbar ^{2N-1}}{\mu ^{3(N-1)}} \frac{1}{H^N} \, . \end{aligned}$$

(176)

This means that we are only interested at the $\hbar ^{2N-1}$ term in the Wigner-Weyl transform since that is the only contributing factor to our desired classical limit. This term is the $s = 0$ term in our calculations. The leading $\hbar$ correction is then ${\mathcal {O}}(\hbar ^{2N+1})$, corresponding to $s=1$.

Since only the $s=0$ terms correspond to the classical limit, then from (173), the recurrence relation that we are interested in studying is

$$\begin{aligned} \alpha ^{(0)}_{m,j} = \frac{1}{m(2N-1+2j)} \sum _{n=1}^m \frac{n a_n}{2^{n-1}} \alpha ^{(0)}_{m-n,j-1} \, , \end{aligned}$$

(177)

where we have let $\alpha ^{(s)}_{m,j}$ for $s > 0$ vanish.

We note in passing that the leading $\hbar$ correction of the time of arrival result is ${\mathcal {O}}(\hbar ^2)$ for nonlinear systems [15]. We then need to let $\hbar$ vanish if we are to recover the correct classical limit for the time of arrival. One could choose $\beta$ such that $\beta \hbar ^{2N-1}$ does not vanish, i.e., $\beta \propto 1/\hbar ^{2N-1}$, so that the leading $\hbar$ correction for $T_C$ becomes ${\mathcal {O}}(\hbar ^2)$. In this scenario, letting ${\mathcal {O}}(\hbar ^2)$ vanish won’t remove the $H^{-N}$ term.

Going back, since $\alpha _{m,2N-1}$ is given by (163), then

$$\begin{aligned} \alpha ^{(0)}_{m,0}= & {} -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \delta _{m,0} \, , \end{aligned}$$

(178)

$$\begin{aligned} \alpha ^{(0)}_{m,1}= & {} -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{1}{2N+1} \frac{a_m}{2^{m-1}} \, , \end{aligned}$$

(179)

$$\begin{aligned} \alpha ^{(0)}_{m,2}= & {} -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{1}{m(2N+3)(2N+1)} \frac{1}{2^{m-2}} \sum _{s=1}^m s a_s a_{m-s} \, , \end{aligned}$$

(180)

$$\begin{aligned} \alpha ^{(0)}_{m,3}= & {} -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{1}{m(2N+5)(2N+3)(2N+1)} \frac{1}{2^{m-3}} \sum _{s=1}^m \frac{s a_s}{m-s} \sum _{r=1}^{m-s} r a_r a_{m-s-r} \, , \end{aligned}$$

(181)

where we have used the recurrence relation (177) to get the other nonvanishing terms. We infer that

$$\begin{aligned} \alpha ^{(0)}_{m,j} = -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{\Gamma \left( N+\frac{1}{2}\right) }{\Gamma \left( N+\frac{1}{2}+j\right) } \frac{1}{2^m} C_{m,j} \, . \end{aligned}$$

(182)

Substituting this to the right side of (177) gives

$$\begin{aligned} \alpha ^{(0)}_{m,j} = -i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \frac{\Gamma \left( N+\frac{1}{2}\right) }{\Gamma \left( N+\frac{1}{2}+j\right) } \frac{1}{2^m} \frac{1}{m} \sum _{s=1}^m s a_s C_{m-s,j-1} \, . \end{aligned}$$

(183)

Comparing (182) and (183), we get a recurrence relation for $C_{m,j}$,

$$\begin{aligned} C_{m,j} = \frac{1}{m} \sum _{s=1}^m s a_s C_{m-s,j-1} \, . \end{aligned}$$

(184)

From (178) and (182), we see that $\alpha ^{(0)}_{m,0} = -i \frac{\beta }{2^{2N-2}} \delta _{m,0} = -i \frac{\beta }{2^{2N-2}} \frac{1}{2^m} C_{m,0}$, so $C_{m,0} = 2^m \delta _{m,0}$, i.e.,

$$\begin{aligned} C_{m,0} = \delta _{m,0} \, . \end{aligned}$$

(185)

We now go back to the time kernel solution, which takes the form

$$\begin{aligned} T(u,v) = T_{\text {TOA}}(u,v) + \sum _{j=0}^\infty \sum _{m=0}^\infty \alpha _{m,2N-1+2j} u^m v^{2N-1+2j} \, . \end{aligned}$$

(186)

From (166), this becomes

$$\begin{aligned} T(u,v) = T_{\text {TOA}}^{(0)}(u,v) + \sum _{j=0}^\infty \sum _{m=0}^\infty \left( \frac{\mu }{2\hbar ^2}\right) ^j \alpha ^{(0)}_{m,j} u^m v^{2N-1+2j} \, , \end{aligned}$$

(187)

wherein we only consider the $s = 0$ term. In the time of arrival solution, only the $s = 0$ term is taken as well [15]. With our condition that $\beta \hbar ^{2N-1}$ does not vanish, we see that this equation is the leading order solution to the general potential case, as we have let ${\mathcal {O}}(\hbar ^2)$ vanish. To continue, we use (182) to get

$$\begin{aligned} T(u,v) = T_{\text {TOA}}^{(0)}(u,v) - i^{2N-1} \frac{\beta }{\mu ^{2N-2}} \sum _{j=0}^\infty \sum _{m=0}^\infty \left( \frac{\mu }{2\hbar ^2}\right) ^j \frac{\Gamma \left( N+\frac{1}{2}\right) }{\Gamma \left( N+\frac{1}{2}+j\right) } \frac{1}{2^m} C_{m,j} u^m v^{2N-1+2j} \, . \end{aligned}$$

(188)

Thus, the Wigner-Weyl transform (20) gives

$$\begin{aligned} \begin{aligned} {\mathcal {T}}_\hbar (q,p)&= t_{\text {TOA}}(q,p) - \frac{\mu }{\hbar } \frac{\beta }{2^{2N-2}} \sum _{j=0}^\infty \sum _{m=0}^\infty \left( \frac{\mu }{2\hbar ^2}\right) ^j \frac{\Gamma \left( N+\frac{1}{2}\right) }{\Gamma \left( N+\frac{1}{2}+j\right) } \frac{1}{2^m} C_{m,j} (2q)^m \\&\qquad \qquad \qquad \quad \times \int _{-\infty }^\infty v^{2N-1+2j} {{\,\mathrm{sgn}\,}}(v) \exp \left( -i\frac{p}{\hbar }v\right) \hbox {d}v \, , \end{aligned} \end{aligned}$$

(189)

which becomes, after using (24) and some simplifications,

$$\begin{aligned} \begin{aligned} {\mathcal {T}}_\hbar (q,p)&= t_{\text {TOA}}(q,p) + \beta \frac{(2N-1)!}{2^{N-1}} \frac{\hbar ^{2N-1}}{\mu ^{3(N-1)}} \\&\qquad \qquad \qquad \quad \times \left( \frac{2\mu }{p^2}\right) ^N \sum _{k=0}^\infty (-1)^k \frac{\Gamma (N+k)}{k!\Gamma (N)} \left( \frac{2\mu }{p^2}\right) ^k k! \sum _{m=0}^\infty C_{m,k} q^m \, . \end{aligned} \end{aligned}$$

(190)

Note that for the general potential (93),

$$\begin{aligned} \frac{1}{H^N} = \left( \frac{2\mu }{p^2}\right) ^N \sum _{k=0}^\infty (-1)^k \frac{\Gamma (N+k)}{k!\Gamma (N)} \left( \frac{2\mu }{p^2}\right) ^k \left( \sum _{s=1}^\infty a_s q^s\right) ^k \, , \end{aligned}$$

(191)

for sufficiently small q. We then are left with showing that

$$\begin{aligned} k! \sum _{m=0}^\infty C_{m,k} q^m = \left( \sum _{s=1}^\infty a_s q^s\right) ^k \, , \end{aligned}$$

(192)

so that the shifting term approaches the correct classical limit.

Firstly, for $k=0$,

$$\begin{aligned} \sum _{m=0}^\infty C_{m,0} q^m = 1 \, , \end{aligned}$$

(193)

and so, by (185), we know that this equality holds. To show that this holds for $k > 1$, we first differentiate both sides of (192) with respect to q,

$$\begin{aligned} k! \sum _{m=0}^\infty m C_{m,k} q^{m-1} = k \left( \sum _{s=1}^\infty a_s q^s\right) ^{k-1} \sum _{r=1}^\infty r a_r q^{r-1} \, . \end{aligned}$$

(194)

Suppose (192) is true; we can rewrite the above equation into

$$\begin{aligned} k! \sum _{m=0}^\infty m C_{m,k} q^{m-1} = k \left( (k-1)! \sum _{n=0}^\infty C_{n,k-1} q^n\right) \sum _{r=1}^\infty r a_r q^{r-1} \, . \end{aligned}$$

(195)

We rewrite the double sum as

$$\begin{aligned} \sum _{m=0}^\infty m C_{m,k} q^{m-1} = \sum _{m=0}^\infty \sum _{j=0}^m (m-j) a_{m-j} C_{j,k-1} q^{m-1} \, , \end{aligned}$$

(196)

or,

$$\begin{aligned} \sum _{m=0}^\infty m C_{m,k} q^{m-1} = \sum _{m=0}^\infty \sum _{s=0}^m s a_s C_{m-s,k-1} q^{m-1} \, , \end{aligned}$$

(197)

and thus, we obtain

$$\begin{aligned} m C_{m,k} = \sum _{s=0}^m s a_s C_{m-s,k-1} \, . \end{aligned}$$

(198)

By (184), we know that this equality holds as well, which implies that (192) is indeed true.