If I tell you all the averages of powers of position and momentum in quantum mechanics can you tell me the value of the wave-function? What can you tell me about the wavefunction? Is there anything about this in the literature?

Attempted Proof

I'm uncertain about the proof (I think the answer is no) and it's validty but here's my attempt. Starting with the modulus of the the wave-function in position space:

$$ |\psi(x,t)|^2 = \psi(x,t)^* \psi(x,t)$$

Let us expand around $z$:

$$ |\psi(x,t)|^2 = \Big( \psi(z,t)^* + (x -z)\psi'(z,t)^* + \frac{(x -z)^{2}}{2!}\psi''(z,t )^* +\dots \Big )$$

$$\cdot \Big (\psi(z,t) + (x -z)\psi'(z,t) + \frac{(x -z)^{2}}{2!}\psi''(z,t)+\dots \big)$$

Multiplying the R.H.S brackets together. We will refer to the below equation as equation $\lambda$:

$$ |\psi(x,t)|^2 = | \psi(z,t)|^2 + (x -z)\psi'(z,t)^* \psi(z,t) + (x -z)\psi'(z,t) \psi(z,t)^* +\dots $$

Let us substitute $x \to z + c/(2N) \int_{-N}^N dz$ in equation $\lambda$:

$$ |\psi(z + \frac{c}{2N} \int_{-N}^N dz,t)|^2 \cdot 1 = | \psi(z,t)|^2 + (\frac{c}{2N} \int_{-N}^N dz)\psi'(z,t)^* \psi(z,t) + (\frac{c}{2N} \int_{-N}^N dz)\psi'(z,t) \psi(z,t)^* +\dots $$

Multiplying $2N$ both sides , rearranging terms and taking the limit $N \to \infty$:

$$ \lim_{N \to \infty} 2N (|\psi(z + \frac{c}{2N} \int_{-N}^N dz,t)|^2 - | \psi(z,t)|^2) = \lim_{N \to \infty} ( c \int_{-N}^N dz)\psi'(z,t)^* \psi(z,t) + ( c \int_{-N}^N dz)\psi'(z,t) \psi(z,t)^* +\dots $$

The R.H.S can be now expressed as a function of momentum averages as it has terms such as $\int_{-\infty}^\infty \psi'(z)\psi(z) dz$. Hence,

$$ \lim_{N \to \infty} 2N (|\psi(z + \frac{c}{2N} \int_{-N}^N dz,t)|^2 - | \psi(z,t)|^2) = f(c,z,\langle p \rangle, \langle p^2 \rangle, \langle p^3 \rangle, \dots) $$

Similarly if we define $\tilde \psi(p,t)$ in momentum space:

$$ \lim_{N \to \infty} 2N (|\tilde \psi(\tilde z + \frac{c}{2N} \int_{-N}^N d \tilde z,t)|^2 - | \tilde \psi(\tilde z,t)|^2) = g(c, \tilde z,\langle x \rangle, \langle x^2 \rangle, \langle x^3 \rangle, \dots) $$

We can compare powers of $c$ and learn something about the wavefunction.

  • $\begingroup$ I would naively understand "averages of powers of a wavefunction" as meaning expectation values of observables over multiple copies of the state. I.e. given $\rho$ you measure $\mathrm{Tr}(A\rho)$ for $A$ observable. To do powers you would then measure $\mathrm{Tr}(B\rho^{\otimes n})$ where $B$ is an observable in the space comprising multiple copies of the state. Is that what you mean here? I don't understand how the calculations you show relate to the stated goal $\endgroup$
    – glS
    Dec 10, 2020 at 8:48
  • $\begingroup$ Yes that is what I mean . The calculations I do relate a Taylor expansion of the wavefunction to the averages. $\endgroup$ Dec 10, 2020 at 9:45
  • $\begingroup$ My method relies more on $\int \psi^* $ observable $ \psi dx $ method for averages rather than the density matrix formulation $\endgroup$ Dec 10, 2020 at 9:47
  • $\begingroup$ ok, but I don't understand why you start doing Taylor approximations for this. If you are asking what you can know about a state given a number of expectation values of observables that's a problem of linear algebra. Setting $\mathrm{Tr}(A\rho)=\alpha$ for $\alpha\in\mathbb R$ and $A$ an observable defines a hyperplane in the space of states. I use density matrix formalism because in the space of density matrices you can understand the constraint as a linear one. Pure states are then found on the boundary of the corresponding set of states. $\endgroup$
    – glS
    Dec 10, 2020 at 12:12
  • $\begingroup$ I start doing Taylor approximations because over there the wavefunction is differentiated which is a reminiscent of momentum. $\endgroup$ Dec 10, 2020 at 12:41

2 Answers 2


The covariance matrix is a function of the expectation values of powers of position and momentum associated to some state in a continuous-variable system. $$ \mathbf{\sigma} = \begin{pmatrix} \langle\hat{x}^2\rangle & \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\} \rangle \\ \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\}\rangle & \langle\hat{p}^2\rangle\end{pmatrix}$$

For states which are fully describable by their first and second moments, like Gaussian States, there is an equivalence between the Wigner Function of the state and its covariance matrix $$W(\mathbf{r}) =\frac{\exp\left( -\frac{1}{2} (\mathbf{r} -\bar{ \mathbf{r}}) \mathbf{\sigma}^{-1} (\mathbf{r}-\bar{\mathbf{r}} ) \right)}{(2 \pi)^N \sqrt{\text{det} \mathbf{\sigma}}}$$ where in this case $\mathbf{r} =(\hat{x},\hat{p})$. So for a given Gaussian State, given some average expectation values, you could generate the average covariance matrix and from that your Wigner Function.

Depending on how many of these "powers" you know, the more statistical moments of your Wavefunction you know allowing you to reconstruct the Wignerfunction of such a wavefunction from its covariance matrix more faithfully in a numerical way.

In Quantum Computing, there is the idea of Quantum State Tomography which is to construct a state from measurements of a statistical ensemble of states so this might be worth looking into also.

  • $\begingroup$ Where can I read up more on the covariance matrix? $\endgroup$ Jan 14, 2021 at 11:07
  • $\begingroup$ Any Introduction to Quantum Optics textbook will introduce the Covariance Matrix . Walls & Milburn is a good one. Alternatively, you might wish to read up on the covariance matrix from a purely mathematical standpoint where it is used to be able to create probability distributions of multiple variable. Here it would be good to start with a two variable Gaussian Distribution. $\endgroup$ Jan 14, 2021 at 11:12

After thinking about it ... I came up with a peculiar answer. Consider the problem in density matrix formulation. We are given:

$$ \langle \hat O^s \rangle = \text{Tr } \rho \hat O^s $$

Let us write in terms of eigenvalues $ O_i$ of $\hat O$:

$$ \langle \hat O^s \rangle = \sum_{i,j =1}^N \rho_{i,j} O_j^s $$

Let us scalar multiply with a constant $k_0$ raise to the power $1/s$ and take limit $s \to \infty$:

$$ \lim_{s \to \infty}\langle k_0 \hat O^s \rangle^{1/s} = \begin{cases} O_{max} & k_0 \sum_{i=1}^N \rho_{i,max} \geq 1 \\ 0 & \text{otherwise} \end{cases} $$

To convince yourself of the above consider the toy version with $s \to \infty$ and $a> b>1$: $$ \lim_{s \to \infty}(a5^s + b3^s)^{1/s} = 5$$

Let us assume we are lucky or finetune $k_0$ and do not get $0$. Now to get the next eigenvalue we can subtract by $ (k_1 O_{max})^s$. Then we will fine tune $k$ such that:

$$ \lim_{s \to \infty} ( \langle k_0 \hat O^s \rangle - (k_0 k_1 O_{max})^s )^{1/s} = \begin{cases} O_{max - 1} & k_0 k_1 \sum_{i=1}^N \rho_{i,max - 1} \geq 1 \text{ and } k_1 = \sum_{i=1}^N \rho_{i,max} \\ 0 & \text{otherwise} \end{cases} $$

Keep repeating until all eigenvalues are found and all summation of columns (out of order) of $\rho$ are found. The columns must be now arranged in a way such that $\text{ Tr} \rho = 1$. I suspect this is the best we can do

  • $\begingroup$ I still don't get the point of this. If the problem is, given averages of observables $\langle \mathcal O_k\rangle_\rho$, to find $\rho$, then it's simply a problem of linear algebra. Each constraint $\langle \mathcal O_k\rangle_\rho=\alpha_k\in\mathbb R$ bounds the state to be in a given hyperplane in state space. The intersection of the constraints, if non-empty, is the set of states compatible with the observations $\endgroup$
    – glS
    Dec 19, 2020 at 9:23
  • $\begingroup$ Maybe I'm missing something. Feel free to post your answer $\endgroup$ Dec 19, 2020 at 9:26
  • $\begingroup$ I guess it mostly depends on the exact question. In fairness, thinking about it some more, if the question is really "given $\langle x^j\rangle_\psi,\langle p^k\rangle_\psi$ can/how do you find $|\psi\rangle$" then I'm not sure what's the answer. I think you can recover $|\psi(x)|^2$ from the expectation values of $\langle e^{ip\hat x}\rangle$ by Fourier transform. Doing the same for the momentum representation of the wavefunctino should give you enough information to reconstruct the state, but I need to think more about it $\endgroup$
    – glS
    Dec 19, 2020 at 19:13
  • $\begingroup$ what you are doing in this answer seems like a convoluted way to write the operator norm of the observable, but I don't understand how that is connected with the original question. The calculation you present looks like Gelfand's formula for the operator norm, but position and momentum observables are unbounded, so that won't work in this case $\endgroup$
    – glS
    Dec 19, 2020 at 19:25

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