Let us take a many-body quantum system, whose phases in the configuration basis are labeled by $\mathbf {\hat q}=(q_1,\cdots, q_N)$ and momenta $\mathbf {\hat p}=\left(-i\frac{\partial}{\partial \hat q_1},\cdots, -i\frac{\partial}{\partial \hat q_N}\right)$. Let us then consider the operator \begin{equation*} f(\mathbf {\hat q}, \mathbf {\hat p})\equiv \hat q_1^{n_1}\cdots \hat q_N^{n_N}\left(-i\frac{\partial}{\partial \hat q_1}\right)^{m_1}\cdots \left(-i\frac{\partial}{\partial \hat q_N}\right)^{m_N} \end{equation*} of powers of configurations and positions, $n_i, m_i\in \mathbb N^0$.

Is it correct that the object \begin{equation*} \tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}\equiv\int_{\mathbb{R}^N} \mathrm d\mathbf {\hat q} \left\langle\mathbf q\middle|f\left(\mathbf {\hat q} ,-i\frac{\partial}{\partial {\mathbf {\hat q} }}\right) \middle|\mathbf q\right\rangle \end{equation*} is NOT defined (i.e. it is not a well posed trace)?

In particular, for infinite-dimensional Hilbert spaces $H$, an operator is trace class if it is bounded. In my case this is not supposed to be the case, as \begin{equation} \sup_{|\mathbf q\rangle\in\mathcal D(H), ||\mathbf r||\neq 0}\frac{||f\left(\mathbf q,-i\frac{\partial}{\partial {\mathbf q}}\right) |\mathbf q\rangle ||}{|| |\mathbf q\rangle ||}=+\infty \end{equation} where $\mathcal D(H)$ is the (unbounded) domain in the Hilbert space of definition of the operator; in particular, $\hat f$ is the product of powers of unbounded operators.

Instead, in case one includes a canonical weight and defines \begin{equation*} \mathrm{tr}\{e^{-\beta\hat H}f(\mathbf {\hat q} ,\mathbf {\hat p} )\}\equiv \int_{\mathbb{R}^N} \mathrm d\mathbf {\hat q} \left\langle{\mathbf{\hat q}} \middle|e^{-\beta\hat H}f\left(\mathbf {\hat q} ,-i\frac{\partial}{\partial {\mathbf {\hat q} }}\right) \middle|\mathbf {\hat q} \right\rangle \end{equation*} is the equation above a well defined trace?

  • $\begingroup$ (1) you are right, this trace is not defined, (2) it is true that trace class operators are bounded, but the converse is false (not even every bounded operator has well defined trace), (3) even for trace class operators, trace is defined as $\mathrm{tr}(A) = \sum_i \langle i | A | i \rangle$, where kets $| i \rangle$ form on orthonormal basis. On the other hand "kets" $| q \rangle$ that you use are not even well defined vectors in the Hilbert space. Concerning (4), this depends on the form of $\widehat H$. $\endgroup$
    – Blazej
    Commented Mar 22, 2021 at 21:24
  • $\begingroup$ This is the celebrated standard ordering (normal ordering, essentially) prescription, or Mehta prescription. Perfectly finite and meaningful for sensible functions. You may try to prove the nonexistence of fish, but tell that to the fishermen on the wharf. $\endgroup$ Commented Mar 23, 2021 at 0:17

1 Answer 1


I'll give it a shot. Since this is a many particle configuration, we have that $[q_i, q_j] = 0$ and $[p_i, p_j]=0$ and $[q_i, p_j] = i \delta_{ij}$ where the indices label the particles. Therefore, the integral will become separable. Namely, we have that

\begin{align*} &\tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}\equiv\int_{\mathbb{R}^N} \mathrm d\mathbf {\hat q} \left\langle\mathbf q\middle|f\left(\mathbf {\hat q} ,-i\frac{\partial}{\partial {\mathbf {\hat q} }}\right) \middle|\mathbf q\right\rangle\\ &=\int dq_1\cdots\int d q_N \langle q_1|\langle q_2| \cdots\langle q_N| \left(\hat q_1^{n_1}\cdots \hat q_N^{n_N}\left(-i\frac{\partial}{\partial \hat q_1}\right)^{m_1}\cdots \left(-i\frac{\partial}{\partial \hat q_N}\right)^{m_N} | q_1 \rangle |q_2\rangle \cdots |q_N\rangle \right) \end{align*}

where I have made use of the definition of the direct product on a hilbert space for $N$ particles.

Given our commutation relations, these integrals are separable. That is,

\begin{align*} &\tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}=\prod_{i=1}^N \int dq_i \langle q_i| \hat{q}_i^{n_i}\hat{p}_i^{m_i}|q_i\rangle \end{align*}

Now this we can deal with. first recall that $\langle q_i| \hat{q}_i = \langle q_i| q_i$ so that

\begin{align*} &\tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}=\prod_{i=1}^N \int dq_i \int dp_i \langle q_i| q_i^{n_i}\hat{p}_i^{m_i}|q_i\rangle \end{align*}

To take care of the momentum we insert unity resolved in the momentum basis of particle $i$ so that

\begin{align*} &\tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}=\prod_{i=1}^N \int dq_i \int dp_i \langle q_i| q_i^{n_i}\hat{p}_i^{m_i}|p_i \rangle \underbrace{\langle p_i|q_i\rangle}_{\frac{e^{-iq_ip_i}}{\sqrt{2\pi}}}\\ &=\frac{1}{(2\pi)^{\frac{N}{2}}}\prod_{i=1}^N \int dq_i \int dp_i q_i^{n_i} p_i^{m_i} \langle q_i|p_i \rangle e^{-ip_i q_i}\\ &=\frac{1}{(2\pi)^{\frac{N}{2}}}\prod_{i=1}^N \int dq_i \int dp_i q_i^{n_i} p_i^{m_i} e^{ip_i q_i} e^{-ip_i q_i}\\ &=\frac{1}{(2\pi)^{\frac{N}{2}}}\prod_{i=1}^N \left(\int_{\mathbb{R}} dq_i q_i^{n_i}\right)\left( \int_{\mathbb{R}} dp_i p_i^{m_i} \right) \end{align*}

so indeed it appears that unless we impose a momentum cutoff and restrict ourself to a finite region of space then what we have is just a big product of divergences, and hence is not a well defined map from $\mathcal{H}\to \mathbb{R}$.

As for your question regarding the weighting factor of the hamiltonian I believe that the answer ought to depend on what the actual hamiltonian is, but if you think that's incorrect I can reconsider and try to approach that for general $\hat{H}$.

  • $\begingroup$ Thanks a lot for the detailed answer. I think by adding a canonical weight $e^{-\beta\hat H}$ within the braket a general evaluation of the trace would break down. In particular we would not be allowed to factorize the average into single particles contributions, apart from very specific cases (classically integrable). Is it correct? $\endgroup$
    – Graz
    Commented Jan 5, 2019 at 10:37
  • $\begingroup$ Yea it would certainly be more involved, but I imagine that depending on the Hamiltonian it could possibly help with convergence.. something like an exponential regulator maybe $\endgroup$ Commented Jan 5, 2019 at 10:40
  • 1
    $\begingroup$ It may not be factorable but.. doable.. I think it will depend .. I think it would be both factorizable And convergent with a SHO Hamiltonian $\endgroup$ Commented Jan 5, 2019 at 10:42
  • $\begingroup$ Ok I see. Anyway, I am not sure that $\tilde{\mathrm{tr}}$ defines a proper trace in the framework of a rigged Hilbert space. In particular the operators within the average are not trace class, as they are not bounded. Maybe imposing a cutoff would solve the boundedness issue, but some basic properties of the trace may be lost (i.e. ciclicity). Do you think it's correct? $\endgroup$
    – Graz
    Commented Jan 5, 2019 at 11:38
  • $\begingroup$ Yea I would agree that $\tilde tr$ is not a true trace operation.. the momentum cutoff wouldn’t be of immediate help (I believe) since the completeness relation wouldn’t naively hold (I believe) $\endgroup$ Commented Jan 5, 2019 at 11:48

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