Given averages of powers of position and momentum in quantum mechanics what information can be secured about the wave-function?

Question

Question

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.

Answer 1

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.

Answer 2

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