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Born's statistical interpretation of the wave function says that $|\Psi (x,t)|^2$ is the probability density of finding the particle at point $x$ at time $t$, then

$$\int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1 \tag{1}$$

In other words, the particle must be somewhere in space at a certain time $t$. Is this equivalent to saying that the particle must be somewhere in time at a certain position $x$?

My Thoughts: Why I can't write,

$$\int_{-\infty}^\infty |\Psi (x,t)|^2 dt = 1 \tag{2}$$

Whenever I look for the position of the particle, $\Psi$ stops obeying the Schrödinger equation and discontinuously collapses to a spike around some position $x$. If I was able to focus my microscope at say $x = 2$ and look for the particle across all of time (in this case, time is the measurement at not position so setting my microscope at $x = 2$ doesn't disturb anything), then $\Psi$ would collapse to a spike around some value of $t$ and this method of normalizing the wave function would be appropriate. However, humans can only sample at instants in time and look over all space (equation $\textbf{(1)}$). We can't sample at an instant in position and look over all time (equation $\textbf{(2)}$). Scientists can't search for the particle in time. Therefore $\textbf{(2)}$ is not appropriate. However, even though we can't do this as humans (search through time at will instead of staying anchored to the present), is it wrong to say that 'nature' can't accomplish $\textbf{(2)}$? Or do we have laws like the 2nd Law of Thermodynamics that say nature is prohibited from doing so? Is this an example that shows how space and time are not on equal footing?

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  • $\begingroup$ "Whenever I look for the position of the particle, Ψ stops obeying the Schrödinger equation and discontinuously collapses to a spike around some position x. " This is false. Time evolution in quantum mechanics is always according to the Schrödinger equation. $\endgroup$ – Robin Ekman Aug 15 '17 at 1:18
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    $\begingroup$ @RobinEkman That's not the way they teach Copenhagen. Of course after the collapse the wave function evolves according to the Schrodinger equation again $\endgroup$ – DWade64 Aug 15 '17 at 2:11
  • $\begingroup$ Well no, but no one should take Copenhagen seriously. $\endgroup$ – Robin Ekman Aug 15 '17 at 16:26
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Time isn't a quantum-mechanical observable; it's a label. To understand the difference we must consider classical mechanics, in which canonical coordinates are functions of a time label. In particular, time doesn't have a conjugate momentum with which it has a canonical Poisson bracket.

Similarly, in field theory the action is a spacetime integral over a function of spacetime-dependent fields and their derivatives. These fields play the role of canonical fields and their arguments play the role of a time variable, so that even space isn't an observable in this context because we measure the field amplitude at a spacetime event, not a single particle's location.

The functional derivative $\frac{\delta\phi (x)}{\delta \phi (y)}=\delta (x-y)$ and Poisson bracket $\{\phi(x),\,\pi(y)\}=\delta(x-y)$ generalise analogous results from quantum mechanics, and show how the labels relate the quantities that become observables when we quantise this theory.

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For a quantum system with one degree of freedom on the closed interval $I$, the Hilbert space is $L^2(I)$. In this case the $I$ is the range for the spatial coordinate $x$, so that normalisation applies with respect to the Lebesgue measure on $I$. Now suppose that you have a dynamics described by the Hamiltonian $H$ on such Hilbert space, and that $\psi$ is an eigenstate of $H$ with eigenvalue $E$. The time evolution of $\psi$ is

$$t\mapsto e^{iEt}\psi.$$

If we naively try to integrate this function, we get to

$$\int_{\mathbb R}|e^{iEt}\psi(x)|^2\text d t = \int_{\mathbb R}|\psi(x)|^2\text d t = |\psi(x)|^2\int_{\mathbb R}\text dt,$$

which is infinite for every $x\in I$ for which $\psi(x)\neq 0$ or zero otherwise. We then have a problem trying to attach the meaning of probability to such integral. You can interpret this result as saying that a particle will pass through $x$ infinitely often provided $\psi(x)\neq 0$ if you wait indefinitely, but this information is already deductible from $\psi$ itself, there is no need to do such an integral.

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  • $\begingroup$ Thank you for the response. You give a nice specific example of an energy eigenstate being non-normalizable with respect to time. To play along with the question a little more, all that your response says is that states of the form $\exp ^{iEt} \psi_{E}(x)$ are not physically realizable states in a world where you can sample across all of time at any given moment. This is a world where probabilities are given between any two points on your timeline (instead of probabilities of finding the particle between any two x's). Is there a more fundamental response you could give? $\endgroup$ – DWade64 Aug 14 '17 at 19:16
  • $\begingroup$ The schrodinger equation uses a single derivative of time and a second derivative of position (unequal footing). Does this force probability densities to be normalized with respect to position? Unless I'm not understanding your response fully, I don't believe that your response prevents normalization with respect to time, but shows an example when it fails $\endgroup$ – DWade64 Aug 14 '17 at 19:16
  • $\begingroup$ Interesting. Actually in this example that you give, and if we are in the world that I'm talking about (normalizing with respect to time), you example shows that you can't measure energy in my world. My world essentially stops the flow of time. Your response shows that if this is the case, there is no such thing as states of definite energy. (Also assuming that if you do normalize with respect to time, the wavefunction stays normalized with respect for position. Else your normalization constant would change with position which might impose extra conditions on the wave function other than S.E.) $\endgroup$ – DWade64 Aug 14 '17 at 19:26
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    $\begingroup$ From a mathematical point of view, the formalism of QM stems from the canonical commutation relations, which involve the position and momentum operators. There is no such a commutation relation for time and energy (it is common belief that one cannot associate an observable operator to time; see for instance physics.stackexchange.com/q/6584). $\endgroup$ – Phoenix87 Aug 14 '17 at 19:52
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    $\begingroup$ I don't see the connection between the possibility of realising a state and the fact that an integral is not finite in this case. From a physical point of view, it makes little sense integrating in time over all of $\mathbb R$. It is more realistic to consider a time lapse (e.g. a time interval), which you can associate to a measurement. Then your integral will be finite and you can normalise any other of such integrals to this value. However, as shown above, such a factor will just cancel out, showing that the information you want is already encoded in the wavefunction. $\endgroup$ – Phoenix87 Aug 14 '17 at 20:48
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Given that $$\int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1,\tag1$$ implies that the particle must be somewhere in space at any time does not imply that it must be somewhere in time at a certain position $x$. To see this consider the simple example of a particle in a one dimensional box (infinite potential well). The wavefunction $\psi$ and the probability density $|\psi|^2$ of the first few eigenstates look like those in the figure bellow

enter image description here

As you can see, there are certain points (nodes) such that $|\psi(t)|^2=0$ meaning that the particle would never be found there.

You can also notice that Eq. (1) represent a sum of probabilities therefore it must be dimensionless. This implies that the dimension of $\psi$ (for one dimensional systems) is $[\mathrm{length}]^{-1/2}$. Now if we assume that (1) and its physical interpretation as a normalization of probabilities are correct then the dimension of $$\int_{-\infty}^\infty |\Psi (x,t)|^2 dt,$$ must be $[\mathrm{time}][\mathrm{length}]^{-1/2}$ which means the above integral cannot be equal to one. It cannot therefore represent a probability.

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    $\begingroup$ While I agree with much of your analysis, we don't define the wavefunction to have a particular unit, rather, we define its integral with respect to whatever to be unitless: if such a wavefunction existed, it would have units of [time]$^{-1/2}$. It's all somewhat pointless as time is not an operator, but your linked answer basically explains that we choose the unit of the wavefunction to satisfy the requirement, not the other way around. $\endgroup$ – user121330 Aug 14 '17 at 22:01
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    $\begingroup$ @user121330 I didn't define the dimension of $\psi$ either. Indeed I just assume that probabilities are dimensionless. $\endgroup$ – Diracology Aug 14 '17 at 22:19
  • $\begingroup$ In that case, I don't understand your last sentence. $\endgroup$ – user121330 Aug 14 '17 at 22:51
  • $\begingroup$ @user121330 I am not quite sure what you mean. I am considering OP's Eqs. (1) and (2). The wavefunction $\psi$ is the same in both equation. Then I assume Eq. (1) represent a normalization of probabilities hence it is dimensionless and $\psi$ has dimension of $[\mathrm{length}]^{-1/2}$. Given this $\psi$, the last step is to verify the dimension of the integral in Eq. (2) and notice that that equation cannot be correct. $\endgroup$ – Diracology Aug 14 '17 at 23:08
  • $\begingroup$ Oh. You're assuming that $|\Psi(x,t)|^2$ has the same units in both equations, probably because it's got the exact same symbols. I'd argue that $\Psi$ has contextual units, if only because that's what every other wavefunction does. You aren't necessarily wrong, but it seems a tad pedantic. On the other hand, I'm talking to you about it, so what does that make me... $\endgroup$ – user121330 Aug 14 '17 at 23:30

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