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Summary: This turned out to be a rather trivial one indeed. As Marek mentioned in the comment, the continuity equation is trivial. And it indeed turns out be so. Godfrey Miller elaborates on this, showing that the continuity equation merely shows that the transition amplitude is constant with time. All this is only confusing as long as you stick to the belief that the continuity equation is some exclusive indication of probability density.

$\rho = \psi_1^* \psi_2$ is the probability density for transition from state 2 to 1. I am having trouble interpreting the probability though $$dP_{1\leftarrow 2}=\psi_1^* \psi_2 d^3 r $$ does this mean the probability in a region of volume $d^3 r$ to go from state 2 to state 1? That does not make sense at all!

EDIT: I more or less agree with Marek's answer. However, that makes it seem this is not the probability density. How do we reconcile that with the fact that it satisfies the continuity equation. If we have two states:
$$ i \hbar \partial_t \psi_{1,2}= -\frac{\hbar^2}{2m} \nabla^2 \psi_{1,2} + V\psi_{1,2}$$

After making standard manipulations (taking conjugate, multiplying by conjugate, subtracting) we would get $$\partial_t ( \psi_1^*\psi_2) +\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)=0$$ This seems consistent with the continuity equation: $$\partial_t \rho +\nabla.\mathbb{j} =0$$ However, we seem to be changing its relation to the probability by making the probability $P=\int \int\rho^*(r,r') \rho(r,r') d^3r d^3r' $ instead of the regular $P=\int \rho(r) d^3r$ For a kind of intuitive "dimensional consistency" can we instead say $P=\int \int \; \sqrt{\rho^* \rho} \; d^3r d^3r'$ $$****$$ Full steps: $\psi_1$ and $\psi_2$ are two states of the same system, i.e both satisfy the SE of the same potential: Taking the SE for $\psi_2$ and multiplying by $\psi_1^*$ : $$ \psi_1^*\times\left( i \hbar \partial_t \psi_2 = -\frac{\hbar^2}{2m} \nabla^2 \psi_2 + V\psi_2\right)$$ Taking the SE for $\psi_1$ complex conjugating the entire equation and multiplying by $\psi_2$ : $$\left( -i \hbar \partial_t \psi_1^*= -\frac{\hbar^2}{2m} \nabla^2 \psi_1^* + V\psi_1^*\right)\times \psi_2$$

($V$ is real, no decay processes etc) On subtracting the second equation from the first

$$\partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \left( \psi_1^* \nabla^2 \psi_2 - (\nabla^2 \psi_1^* )\psi_2\right)$$ which is the same as $$ \partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)$$

yayu
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  • @yayu: I don't think the math leading to $i\hbar\partial_t(\psi_1^*\psi_2) + \cdots = 0$ is valid. Are you sure about that? – David Z Apr 18 '11 at 02:46
  • @David Ok. I will work out the steps for full scrutiny in another edit. – yayu Apr 18 '11 at 02:47
  • Ah, I missed a sign when I was doing it... never mind. (Though you do seem to have an extra $i$ in the denominator of $\hbar^2/2mi$) – David Z Apr 18 '11 at 03:08
  • @David yup.. done! – yayu Apr 18 '11 at 03:26
  • And never mind again, I missed that you divided through by $i\hbar$ :-P – David Z Apr 18 '11 at 03:48
  • @David actually i corrected that factor after seeing your comment. I was copy pasting my intitial equation's latex to be quicker, and I had left the $i\hbar$ factor sticking to the $\partial_t(\cdots)$ so thanks :-P – yayu Apr 18 '11 at 03:50

4 Answers4

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That's probably (no pun intended) because it's not a probability density at all. Wave functions by themselves don't have any physical interpretation. Only their squares do.

The probability of transition is computed like this $$P_{1 \to 2} = {\left \Vert \left < \psi_1 | \psi_2 \right > \right \Vert^2 \over \left \Vert \psi_1 \right \Vert^2 \left \Vert \psi_2 \right \Vert^2}$$

What you've written is just a density of the probability amplitude $\left<\psi_1|x\right>\left<x|\psi_2\right>$. It can be complex so it certainly cannot be probability in any sense.

Edit regarding your additional comments: assuming that $\left \Vert \psi_1 \right \Vert^2 = \left \Vert \psi_2 \right \Vert^2 = 1$ you could write $P_{1 \to 2}$ like this $$P_{1 \to 2} = {\rm Tr} \left( \left|\psi_1\left> \right<\psi_1\right| \left|\psi_2\left> \right<\psi_2\right| \right) = \int \psi_1(y) \psi_1^*(x) \psi_2(x) \psi_2^*(y) {\rm d} x {\rm d} y$$ interpreting it as a trace of the product of projection operators. But this integral representation is basically never done, it's much easier to only compute the amplitude and square it when you're finished.

As a moral, you shouldn't insist on restricting your attention to some portion of position space. Quantum theory "doesn't like it" (except for few special constructions).

Marek
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    I don't know what you mean about not having the right units: the quantity in the question is indeed dimensionless, as a probability should be. You're right about the rest, especially the fact that this quantity is complex and so can't be a probability, which is the important thing. – Ted Bunn Apr 14 '11 at 14:15
  • @Ted: oh right, thanks. Probably just temporary loss of senses :) – Marek Apr 15 '11 at 03:16
  • @Marek thanks. one further question, does it make sense to say that $dP_{1\rightarrow 2} = \int (\psi_1^* \psi_2)^* \psi_1^* \psi_2 d^3r$... what does a probability of transition mean in a small volume element. Because states usually means a specifications of the values of $\psi$ in all points in space.. then how can we have a probability of state transition in a small volume?? – yayu Apr 15 '11 at 17:12
  • @yayu: no, it doesn't make any sense. Look, states in quantum theory can be extended in $x$ coordinate. When one state transitions into another the contribution to the probability of this process is from the correlation between all parts of the wavefunctions (i.e. correlations between $\psi_1(x)$ and $\psi_2(y)$ with $x \neq y$). You obtain only probability amplitude by a single integral. To get a genuine probability, you need a double interal (i.e. a square, as usual). – Marek Apr 15 '11 at 23:24
  • @marek then what would be an integral representation of $P_{1 \to 2} = {\left \Vert \left < \psi_1 | \psi_2 \right > \right \Vert^2 \over \left \Vert \psi_1 \right \Vert^2 \left \Vert \psi_2 \right \Vert^2}$ – yayu Apr 16 '11 at 03:59
  • I get it. Thanks. I was confused because I tried to technically interpret the word correlation. – yayu Apr 16 '11 at 20:21
  • @Marek sorry for dragging you back to this concerning a retrospective doubt. But i noticed that the quantity $\rho$ appeared to satisfy the continuity equation, and the interpretation says that it must be the probability density, albeit some strange kind of probability density. – yayu Apr 18 '11 at 02:31
  • @yayu: continuity equation in quantum mechanics is always trivial in some sense. Say you have one particle. Then it's obvious that the probability that it is somewhere will always have to sum to unity. Continuity equation just restates this locally. If you have more particles, or more quantum states then obviously the local law will change. You can then interpret $\rho$ as probability density on some space (which will generally be higher dimensional) but its relation to the original position space is not apparent. – Marek Apr 18 '11 at 05:39
  • @Marek: -1 (sorry--- your answer is wrong and upvoted) this is explained properly below in Kalitvianski's answer--- the quantity in question is the probability, knowing that the particle is in state $\psi_1$ that it will be in state $\psi_2$. This is called the "overlap" of the two states. – Ron Maimon Aug 31 '11 at 03:24
  • @Ron: Kalitvianski's answer doesn't really explain anything and I have no idea what you have in mind either. The quantity in question can certainly be interpreted as overlap of wavefunctions but that's simply not a probability. It can be complex for goodness's sake... – Marek Sep 03 '11 at 22:20
  • @Marek: sorry, you are absolutely right. The quantity is the probability amplitude for being in state 2 when you know you are in state 1. States in quantum mechanics are no mutually exclusive, as you know. – Ron Maimon Sep 04 '11 at 21:11
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Since \begin{equation} \langle \psi_1 \mid \psi_2 \rangle = \int d^3 x \ \psi_1^*(t,x) \psi_2(t,x), \end{equation} the quantity $\psi_1^* \psi_2$ is the spatial density of the transition amplitude from $\psi_2$ to $\psi_1$. This amplitude density is in general complex, so it is not possible in general to interpret it as a probability density, but according to your equation it is conserved all the same. The only exception is when $\mid \psi_1 \rangle = \mid \psi_2 \rangle = \mid \psi \rangle$, in which case the equation above becomes \begin{equation} 1 = \int d^3 x \ \psi^*(t,x) \psi(t,x), \end{equation} which is why the non-negative quantity $\psi^* \psi$ can be interpreted as a probability density.

The conservation of amplitude density implies the important fact that transition amplitudes between well-behaved solutions of the Schrodinger equation are time independent. It follows from your equation that \begin{equation} \frac{\partial}{\partial t} \langle \psi_1 \mid \psi_2 \rangle = \int dx \ \frac{\partial}{\partial t} (\psi_1^* \psi_2) = \frac{i \hbar}{2 m} \int dx \ \nabla \cdot (\psi_1^* \nabla \psi_2 - \psi_2 \nabla \psi_1^*). \end{equation} If the wavefunctions $\psi_1$, $\psi_2$ are well-behaved at infinity, this final integral will vanish, so $\langle \psi_1 \mid \psi_2 \rangle = {\rm constant}$. In hindsight, you could have inferred the necessity of the conservation of amplitude density from two facts: 1) transition amplitudes can be written as an integral of a local quantity, and 2) transition amplitudes are time-independent.

Does this help answer your question?

Godfrey Miller
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  • One clarification needed though, what do you mean by: " transition amplitudes can be written as an integral of a local quantity". – yayu Apr 25 '11 at 03:08
  • If a constant quantity can be written as a sum of terms, one term can only change if the other terms change in the opposite sense to compensate. If a constant quantity can be written as an integral over a function of space, that function can only change at one point if it changes elsewhere in the opposite sense to compensate. 2) The Schrodinger equation says that $\partial_t \psi$ only depends on $\psi$ and $\nabla^2 \psi$, so changes in $\psi(x)$ only depend on the values of $\psi$ in the immediate neighborhood of $x$: $\psi$ is therefore a "local" quantity.
    • – Godfrey Miller Apr 26 '11 at 03:38
  • Changes in the quantity $\psi^_1(x) \psi_2(x)$ depend only on the values of $\psi_1$ and $\psi_2$ in an immediate neighborhood of $x$, so it too is a local quantity. The transition amplitude $\langle \psi_1 \mid \psi_2 \rangle$ is therefore the integral of a local quantity. 4) Since $\langle \psi_1 \mid \psi_2 \rangle$ is constant, changes in $\psi^_1 \psi_2$ at a point $x$ must be compensated by opposing changes at other points. Since $\psi^*_1(x) \psi_2(x)$ is a local quantity, those changes only take place in an immediate neighborhood of of $x$.
    • – Godfrey Miller Apr 26 '11 at 03:42
  • Since changes in $\psi_1^* \psi_2$ at a point $x$ are compensated by equal and opposite changes in $\psi_1^* \psi_2$ in the neighborhood of $x$, the quantity $\psi_1^* \psi_2$ is locally conserved, and therefore obeys a continuity equation. Does that clarify what I meant? – Godfrey Miller Apr 26 '11 at 03:46