In combinatorial mathematics and probability theory, the Schrödinger method, named after the Austrian physicist Erwin Schrödinger, is used to solve some problems of distribution and occupancy.

Suppose

:<math>X_1,\dots,X_n \, </math>

are independent random variables that are uniformly distributed on the interval&nbsp;[0,&nbsp;1]. Let

:<math>X_{(1)},\dots,X_{(n)} \, </math>

be the corresponding order statistics, i.e., the result of sorting these n random variables into increasing order. We seek the probability of some event A defined in terms of these order statistics. For example, we might seek the probability that in a certain seven-day period there were at most two days in on which only one phone call was received, given that the number of phone calls during that time was&nbsp;20. This assumes uniform distribution of arrival times.

The Schrödinger method begins by assigning a Poisson distribution with expected value λt to the number of observations in the interval [0,&nbsp;t], the number of observations in non-overlapping subintervals being independent (see Poisson process). The number N of observations is Poisson-distributed with expected value&nbsp;λ. Then we rely on the fact that the conditional probability

:<math>P(A\mid N=n) \, </math>

does not depend on λ (in the language of statisticians, N is a sufficient statistic for this parametrized family of probability distributions for the order statistics). We proceed as follows:

:<math>P_\lambda(A)=\sum_{n=0}^\infty P(A\mid N=n)P(N=n)=\sum_{n=0}^\infty P(A\mid N=n){\lambda^n e^{-\lambda} \over n!},</math>

so that

:<math>e^{\lambda}\,P_\lambda(A)=\sum_{n=0}^\infty P(A\mid N=n){\lambda^n \over n!}.</math>

Now the lack of dependence of P(A&nbsp;|&nbsp;N&nbsp;=&nbsp;n) upon λ entails that the last sum displayed above is a power series in λ and P(A&nbsp;|&nbsp;N&nbsp;=&nbsp;n) is the value of its nth derivative at λ&nbsp;=&nbsp;0, i.e.,

:<math>P(A\mid N=n) = \left[{d^n \over d\lambda^n}\left(e^\lambda\, P_\lambda(A)\right)\right]_{\lambda=0}.</math>

For this method to be of any use in finding P(A&nbsp;|&nbsp;N&nbsp;=n), must be possible to find P<sub>λ</sub>(A) more directly than P(A&nbsp;|&nbsp;N&nbsp;=&nbsp;n). What makes that possible is the independence of the numbers of arrivals in non-overlapping subintervals.