I'm not sure if this question even makes sense in a quantum mechanical context, but I was wondering how does one compute the observed momentum of a specific quantum particle in a quantum many-body system?

What about the kinetic energy and how does it relate to the kinetic energy of the entire system?


1 Answer 1


$\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$The state of a quantum $n$-body system is a sum of tensor products of states of the individual particles: $$\ket\psi=\sum_i\ket{\phi_{i1}}\otimes\cdots\otimes\ket{\phi_{in}}$$

Where each $\ket{\phi_{ij}}\in\mathcal H_j$ and $\mathcal H_j$ is the Hilbert space describing the $j$th particle. The state of the whole system $\ket\psi\in\mathcal H$ lives in the Hilbert space $\mathcal H=\mathcal H_1\otimes\cdots\otimes\mathcal H_n.$ In the example of two particles in 1D space, $\mathcal H_1=\mathcal H_2=L^2(\mathbb R),\mathcal H=L^2(\mathbb R^2),$ and the wavefunction $\psi(x,y)$ of the two-body system decomposes as $\psi(x,y)=\sum_i\phi_{i1}(x)\phi_{i2}(y),$ where the $\phi_{i1}(x)$ are wavefunctions for the first particle and similarly for $\phi_{i2}(y).$ You could also choose a different basis for either particle; e.g. you could write all the $\phi_{i2}$ in the momentum basis, and then the combined wavefunction would be $\psi(x,p_y).$

Any operator on a single-particle Hilbert space, say $A:\mathcal H_k\to\mathcal H_k$ can be lifted to act on the whole of $\mathcal H$ by tensoring it with a bunch of identity operators, one for each of the other Hilbert spaces: $\tilde A=\mathbb I\otimes\cdots\otimes A\otimes\cdots\otimes\mathbb I.$ The lifted operator $\tilde A$ acts on a tensor product by applying $A$ to the appropriate factor. $$\tilde A\ket\psi=\sum_i\ket{\phi_{i1}}\otimes\cdots\otimes A\ket{\psi_{ik}}\otimes\cdots\otimes\ket{\phi_{in}}$$

In your case, if you can construct a momentum observable for your single-particle space, then you can lift that into an observable that acts on the many-body space and measures the momentum of the specified particle. In the two-body example, the momentum operator for the first particle is $\hat p_1=-i\hbar\frac{d}{dx}.$ Lifting it and applying it to the two-particle state gives $\hat p_1\psi(x,y)=\sum_i(\hat p_1\phi_{i1}(x))\phi_{i2}(y)=-i\hbar\sum_i\left(\frac{d\phi_{i1}(x)}{dx}\right)\phi_{i2}(y)=-i\hbar\frac{\partial\psi(x,y)}{\partial x}$ (it is common to use the same symbol for the single-particle and lifted operators). The way you compute e.g. expectation values is the same as always, just with a slightly different expansion: $$\langle\hat p_1\rangle=\bra\psi\hat p_1\ket\psi=-i\hbar\iint_{\mathbb R^2}\psi^*(x,y)\frac\partial{\partial x}\psi(x,y)\;dx\,dy.$$

You can do the same thing with the kinetic energy operators $\hat T_k=\frac1{2m_k}\hat p_k^2.$ They lift to the many-body Hilbert space without fuss and let you extract the kinetic energy associated to any one particle. The kinetic energy operator of the whole system can be defined as the sum of the kinetic energy operators: $\hat T=\hat T_1+\cdots+\hat T_n.$

If you would further like to compute the whole probability distribution of an observable, you can start with an eigenbasis $A\ket{a_\ell}=a_\ell\ket{a_\ell}$ for the single-particle operator. Then you can construct a projection operator $\hat P=\ket{a_\ell}\bra{a_\ell}$ such that $\langle\hat P\rangle=\bra\psi\hat P\ket\psi$ gives the probability (density) of measuring the observable $A$ to be $a_\ell$ (in the single-particle Hilbert space). Finally, you can lift $\hat P$ to the many-body Hilbert space and use it to compute probabilities there, too.

  • $\begingroup$ You nowhere mention that all quantum mechanical calculations are probabilistic, that one needs many measurement to check the calculation. One cannot predict the momentum of a single particle, only the probability that the momentum has a given value. $\endgroup$
    – anna v
    Mar 20, 2022 at 17:01
  • $\begingroup$ Very comprehensive, thank you. $\endgroup$ Mar 20, 2022 at 23:07

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