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- , (en)
- , and (en)
- Position and momentum probability densities for an initial Gaussian distribution. From top to bottom, the animations show the cases (en)
- . Note the tradeoff between the widths of the distributions. (en)
- Ω = 2ω (en)
- Ω = ω (en)
- Ω = ω/2 (en)
- Propagation of de Broglie waves in 1d—real part of the complex amplitude is blue, imaginary part is green. The probability of finding the particle at a given point x is spread out like a waveform, there is no definite position of the particle. As the amplitude increases above zero the curvature reverses sign, so the amplitude begins to decrease again, and vice versa—the result is an alternating amplitude: a wave. (en)
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| dbp:proof
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- The derivation shown here incorporates and builds off of those shown in Robertson, Schrödinger and standard textbooks such as Griffiths. For any Hermitian operator , based upon the definition of variance, we have
we let and thus
Similarly, for any other Hermitian operator in the same state
for
The product of the two deviations can thus be expressed as
In order to relate the two vectors and , we use the Cauchy–Schwarz inequality which is defined as
and thus Equation can be written as
Since is in general a complex number, we use the fact that the modulus squared of any complex number is defined as , where is the complex conjugate of . The modulus squared can also be expressed as
we let and and substitute these into the equation above to get
The inner product is written out explicitly as
and using the fact that and are Hermitian operators, we find
Similarly it can be shown that
Thus, we have
and
We now substitute the above two equations above back into Eq. and get
Substituting the above into Equation we get the Schrödinger uncertainty relation
This proof has an issue related to the domains of the operators involved. For the proof to make sense, the vector has to be in the domain of the unbounded operator , which is not always the case. In fact, the Robertson uncertainty relation is false if is an angle variable and is the derivative with respect to this variable. In this example, the commutator is a nonzero constant—just as in the Heisenberg uncertainty relation—and yet there are states where the product of the uncertainties is zero. This issue can be overcome by using a variational method for the proof, or by working with an exponentiated version of the canonical commutation relations.
Note that in the general form of the Robertson–Schrödinger uncertainty relation, there is no need to assume that the operators and are self-adjoint operators. It suffices to assume that they are merely symmetric operators. (en)
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