Question

Quite generally, quantum states are defined to be positive, trace-class linear maps with trace equal to one on a complex separable Hilbert space H. If we require that these trace-class linear maps be bounded, then they are linear functionals on B(H) in particular that satisfy the properties above. However, I found an example of a trace-class linear map A, with domain D_A=H, that is unbounded, and I find this interesting. In the mathematical literature, only bounded trace-class linear operators are discussed. Is there a particular reason for that?

Answer 1

In quantum mechanics, bounded trace-class linear operators are favored because they align with physical observables and yield finite expectation values, facilitating the understanding of quantized energy levels in atoms and molecular vibrations.

Step-by-step explanation:

The discussion of trace-class linear operators in quantum mechanics typically centers on those that are bounded because they are easier to handle mathematically and describe physical states that we can observe and measure. Bounded operators on Hilbert spaces have well-understood mathematical properties and correspond to physical quantities that are measurable in experiments. In quantum mechanics, a state is represented by a wave function, and the boundedness of operators ensures that physical observables, such as energy or momentum, derived from the wave function, have finite expectation values.

Moreover, states in quantum mechanics are often described as quantized. This concept is crucial for understanding many phenomena, including energy levels in atoms and the vibrations of molecules. Quantum states lead to the quantization of energy and other physical properties, which are not continuous but discrete. This discreteness emerges naturally for bound systems and has been confirmed by countless experiments. As systems increase in size, quantization becomes less apparent, but it remains a fundamental aspect of quantum theory at the microscopic level.