I was talking to a family friend in the field of optics at a quantum scale (not sure the proper name for this) and he was explaining to me why you can only determine either the velocity or location of a particle but not the other. He watered it down for me and told me this: You have an initial equation describing the particle (maybe the Schrodinger equation, but I'm not sure) and the Fourier transform. He said the Fourier transform is an operation. You can use the Fourier transform turn the equation into a new one that describes the velocity precisely, but the position less precise, or you can use it to find the location precisely, but the velocity imprecisely. He said it is irreversible so once you do the transform, you can't go back. My question is: why can't I copy the initial equation down on another piece of paper and give it to somebody else to find the velocity while I determine the location?
[Physics] Why can you only measure velocity or location in a particle
fourier transformheisenberg-uncertainty-principlequantum-information
Related Solutions
The Fourier transform of the wavefunction as a function of q is the wavefunction as a functon of x. The Fourier transform of q(t) is a different thing altogether, and it quantum mechanically corresponds to the matrix elements of the q operator between adjacent energy states.
When the motion is periodic, the Fourier transform over all time is a series of delta function spikes at the locations $2\pi n\over T$ where n is an integer, and T is the classical period. The Fourier series coefficients C_n are the coefficients of these delta-spikes.
The quantum mechanical analogs of $C_n$ for the motion with period T are the matrix elements of the q operator near the diagonal in the energy representation, for the value of N which corresponds to the classical motion
$$ q_{N,N+n} = C_n $$
valid in the limit where $n<<N$. The value of N is J/h where J is the action of the classical orbit.
This correspondence was Heisenberg's central idea in 1925, and he built modern quantum mechanics around this. You can derive it from modern quantum mechanics easily, although this is historically backwards.
If you have a q(t), this must be a wavepacket consisting of a superpositon of a gazillion nearby energy levels close to the energy level corresponding to the classical orbit energy E. From the Heisenberg equation of motion, the matrix elements of q(t) are periodic with periods E_n-E, which is $(n-m)2\pi/T$ by the correspondence principle.
If the state $|\psi\rangle$ is a superposition of many energy levels:
$$ |\psi\rangle = \sum_k C_k |E_k> $$
With all relevant $E_i$ only infinitesimally different from $E_N = E$, If you look at the expected value of q as a function of time in this smeared out state, you get
$$\langle \psi | q(t) |\psi \rangle = \sum_k C_n^* C_{n-k} \langle N| q |N-k\rangle e^{i(E_N - E_{N-k})t} $$
Which, using the correspondence rule for the energy level spacings at large quantum numbers $(E_N - E_{N-k}) = {2\pi k \over T}$, gives the classical Fourier series coefficients of q(t):
$$ q(t) = \sum_n e^{i {2\pi n\over T }} \langle N|q_|N-k\rangle \sum_k C_k^* C_{k-n} $$
When the C's are slowly varying functions of n, so that the superposition does not involve macroscopically separate positions, the sum on the inside is essentially constant, it is independent of n, so that the Fourier series coefficients are identifiable as the matrix elements of q.
In words, smearing the wavefunction over many nearby energy levels tells you that the off-diagonal periodic quantity has a Fourier coefficient proportional to the matrix element in the semiclassical approximation. The natural deductive path historically went the other way, this was how Heisenberg interpreted the matrix elements in the energy basis, as generalizations of Fourier coefficients of classical quantities in time.
So, in particular, the Fourier transform of the saw-tooth form gives you the matrix elements for $\langle E_n|q|E_m\rangle$ for the square-well energy levels for large quantum numbers.
$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk$$ where $\phi(k)$ is eigenfunction of Schrodinger Equation for free particle.(I think)
No, you misunderstood your text book. $\phi(k)$ is a completely arbitrary function. For any function $\phi(k)$ this $\Psi(x,t)$ will be a solution of Schrödinger's equation $$i\hbar\frac{\partial}{\partial t} \Psi(x,t)= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t)$$ just because $$e^{i(kx-\frac{\hbar k^2}{2m}t)}$$ for any $k$ is a solution of Schrödinger's equation.
When you take the above solution $\Psi(x,t)$ at time $t=0$, then you have $$\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{ikx}dk$$
This means that $\phi(k)$ is the Fourier transform of $\Psi(x,0)$. You can invert this transformation and get $$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0) e^{-ikx}dx$$
Best Answer
The no cloning theorem (which follows directly from linearity of time evolution) says you can't make a copy so you can't do that.
But that is not the point. The point is that you can make a state and even of you made a thousands states just like, the state itself will not give good results for position and good results for velocity.
There are states that give good results for position and states that give good results for velocity. But one that is really good at one is bad at the other.
So when you pick a state to have lots of you can pick one that is good at position or pick one that is good at velocity.
So why can a state be good at one and not the other?
First, keep in mind that when you get a position or a velocity you generally change the state. And position and velocity are things where the order you do them always matters. So the states you turn something into when you get position must be different than the thing you turn it into when you velocity (otherwise if it was a state that was a final state for each then neither would change it and the order wouldn't matter for that state but for positions and momentum the order always matters). So this indicates that when you measure position and then velocity and so on that you are getting different results. So a good position gives in good (can be different) results for velocity and vice versa.
So you might then ask why position and velocity have to have the order matter. Some people actually say they use that fact as the basis for getting all of quantum mechanical (technically it isn't velocity it is momentum they use and even more technically it is canonical momentum they use, not kinetic momentum). So some people will stop there.
But they aren't being totally honest. That would be valid if they got everything from that, but they do actually have to generalize it to get everything they have to generalize it to a general Poisson brackets go to an imaginary times a commutator. And just one of those is enough. Having the time evolution is enough to explain everything that happens in time.
So a true and proper explanation would have to explain how you get the results over time. Bit it doesn't end up being different, over time you get the initial state to split as the different options become entangled with the different states of the measurement device and they do it with sizes of the branches that lead to certain frequencies of results as measured by separate devices that measure frequencies.
And then the frequencies follow the same rules that position and momentum can have a small spread of values for one of them or the other one can have a small spread of values. And since the order you do them (now a dynamical consequence of the dynamics of how you actually measure them experimentally) matters for every state then again this is forced upon us. And if you want to compute the spread of values you get from repeated interactions, then you can find the spread of one is given by the standard deviation of a function and the spread of the other is given by the standard deviation of the Fourier transform of the first function
But those equations are just equations, nature doesn't pass along copies of the equations it just dynamically evolves.