Dear user1602, yes, $\psi(x)$ and $\tilde\psi(p)$ are Fourier transforms of one another. This answers the only real question you have asked. So if one knows the exact wave function as a function of position, one also knows the wave function as a function of momentum, and vice versa.

In particular, there is no "wave function" that would depend both on $x$ and $p$. Indeed, such a "wave function" would violate a basic principle of quantum mechanics, the uncertainty principle.

The wave function - that only depends on $x$ or that only depends on $p$ - remembers everything that a particle can and needs to remember about its position and momentum. For example, a good wave function describing a particle localized around $x_0$ and moving with momentum around $p_0$ is given by
$$\psi_{x_0,p_0}(x) = C \exp\left(-K(x-x_0)^2 + ip_0 x/\hbar\right)$$
The constant $K$ determines the width but you see that because of the quadratic term, the wave function is only non-vanishing near $x_0$. On the other hand, the $ipx$ term guarantees that the particle is moving to the right with the right momentum. It's all encoded in the changing phase of the wave function. The more quickly the phase of $\psi(x)$ changes with $x$, the higher is the momentum of the particle. If the phase rotates clockwise or counter-clockwise, the particle is moving to the right or to the left, respectively.

The Fourier transform of the wave function above is something like
$$\tilde \psi_{x_0,p_0}(p) = C' \exp\left(-(p-p_0)^2/K' - ip x_0/\hbar\right)$$
Just try it. SchrÃ¶dinger's equation will guarantee that the wave packet is moving in the right direction - and by the right speed - encoded in $p_0$, and the center-of-mass position of the packet will change accordingly, too. The normalization constants $C,C'$ are physically irrelevant but may be chosen to normalize the state vectors to unity. The parameters $K,K'$ specifying the width are equal, up to a multiplication by a numerical constant and a power of $\hbar$: but it's true that the width in the $x$ representation is inverse to the width in the $p$ representation. That's implied by the uncertainty principle, too.

It is not true that one needs "wave functions" that would depend both on position and momentum. It's the whole point of the uncertainty principle that you may only specify the amplitudes with respect to one of these quantities - the other one doesn't commute with it. If one chooses $\psi(x)$, the position operator is a multiplication by $x$ and the momentum $p$ is simply the operator $-i\hbar\partial/\partial x$. Similarly, for $\tilde\psi(p)$, the momentum operator is the multiplication by $p$ and the position operator $x$ equals $+i\hbar \partial/\partial p$. It's pretty much symmetric with respect to $x,p$.

$i\hbar$ is simply a number, or if you must regard it as an operator, a multiple of the identity. So $\langle i\hbar \rangle=i\hbar$, and so is $\langle -i\hbar \rangle$.

By the way, anticommutator of $\hat{x}$ and $\hat{p}$ is not $[\hat{p},\hat{x}]$, but $\{\hat{x},\hat{p}\}=\hat{x}\hat{p}+\hat{p}\hat{x}$.

## Best Answer

$$ [\hat{x}(t),\hat{p}(t)] = [\hat{x}_0,\hat{p}_0] = i\hbar $$

This is easily proven as follows. We have the unitary time evolution operator given by $\hat{T}(t)$. Unitary means $\hat{T}(t)\hat{T}^{\dagger}(t) = \hat{T}^{\dagger}(t)\hat{T}(t) = \hat{I}$ where $\hat{I}$ is the identity operator.

Let $A_0$ and $B_0$ be arbitrary operators with $[A_0,B_0]=C_0$.

In the Heisenberg picture we have

\begin{align} \hat{A}(t) &= \hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\\ \hat{B}(t) &= \hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t)\\ \hat{C}(t) &= \hat{T}^{\dagger}(t)\hat{C}_0\hat{T}(t)\\ \end{align}

So

\begin{align} [\hat{A}(t),\hat{B}(t)] &= \hat{A}(t)\hat{B}(t) - \hat{B}(t)\hat{A}(t)\\ &= \hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t) - \hat{T}^{\dagger}(t)\hat{B}_0\hat{T}(t)\hat{T}^{\dagger}(t)\hat{A}_0\hat{T}(t)\\ &= \hat{T}^{\dagger}(t)[\hat{A}_0,\hat{B}_0]\hat{T}(t) = \hat{T}^{\dagger}(t)C_0\hat{T}(t)\\ &=C(t) \end{align}

So we see that commutation relations are preserved by the transformation into the Heisenberg picture. Actually, we see that commutation relations are preserved by any unitary transformation which is implemented by conjugating the operators by a unitary operator.

Note that unequal time commutation relations may vary. For example, if we have a Harmonic oscillator Hamiltonian

$$ \hat{H}_0 = \frac{1}{2}m\omega^2 \hat{x}^2 + \frac{1}{2m} \hat{p}^2 $$

We can derive and solve the Heisenberg equations of motion. Upon doing this we will get the usual classical equations of motion but with hats.

\begin{align} \hat{x}(t) = \hat{x}_0 \cos(\omega t) + \hat{p}_0 \sin(\omega t)\\ \hat{p}(t) = \hat{p}_0 \cos(\omega t) - \hat{x}_0 \sin(\omega t) \end{align}

Consider $\hat{x}(0) = \hat{x}_0$ and $\hat{x}(\frac{\pi}{2}\frac{1}{\omega}) = \hat{p}_0$.

\begin{align} \left[\hat{x}(0),\hat{x}\left(\frac{\pi}{2}\frac{1}{\omega}\right)\right] = [\hat{x}_0,\hat{p}_0] = i\hbar \end{align}

So we see that the commutator of $\hat{x}$ with itself at different times can be non-zero.