Hilbert Space and Tensor Product Questions.

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SUMMARY

The discussion centers on the relationship between Hilbert spaces and tensor products in quantum mechanics, specifically regarding the state ∣E,l,m⟩. It is established that this state can represent either a linear combination of individual states for a single particle or a tensor product for a composite system. The ability to split a Hamiltonian into separate parts for independent solutions is contingent on the system's degrees of freedom, with specific examples provided for one and two-particle systems. The Hilbert space utilized depends on the Hamiltonian's form, such as square-integrable functions for a particle in a box.

PREREQUISITES
  • Understanding of Hilbert spaces in quantum mechanics
  • Familiarity with Hamiltonians and their role in quantum systems
  • Knowledge of tensor products and their application in composite systems
  • Concept of eigenfunctions and their completeness in relation to Hilbert spaces
NEXT STEPS
  • Study the properties of tensor products in quantum mechanics
  • Learn about the role of Hamiltonians in defining Hilbert spaces
  • Explore the concept of eigenfunction completeness in various quantum systems
  • Investigate the implications of particle interactions on Hamiltonian separability
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Quantum physicists, students of quantum mechanics, and researchers exploring the mathematical foundations of quantum systems will benefit from this discussion.

lackrange
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I never thought about this stuff much before, but I am getting confused by a couple of things.
For example, would the state ∣E,l,m⟩ be the tensor product of ∣E⟩, ∣l⟩, ∣m⟩, ie. ∣E⟩∣l⟩∣m⟩? I always just looked at this as a way to keep track of operators that had simultaneous eigenfunctions in a Hilbert Space, but now I am confused by the lingo. If it is a tensor product, then is there a difference between their Hilbert spaces? Given a Hamiltonian, when can you split it up into two and solve them separately and then take the tensor product of the solutions to call it the solution to the total Hamiltonian?

Also, when given a Hamiltonian, how do you know what Hilbert space you are working in? I know wave functions are square integrable, but for instance in the infinite well problem, the eigenfunctions vanish outside the well, so when we say eigenfunctions of an observable are complete, with respect to which Hilbert space? (I know in the infinite well problem it is the space of functions that vanish outside the well, but how do you know in general what your eigenfunctions span?)
 
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The state ∣E,l,m⟩ is not necessarily the tensor product of ∣E⟩, ∣l⟩, and ∣m⟩. It depends on whether you are describing a single particle or a composite system. If you are describing a single particle, then the state would be a linear combination of the individual states ∣E⟩, ∣l⟩, and ∣m⟩. If you are describing a composite system, then the state could indeed be a tensor product of the individual states. In general, when given a Hamiltonian, the Hilbert space you work in depends on the form of the Hamiltonian. For instance, if the Hamiltonian is describing a particle in a box, then the Hilbert space would be the space of square-integrable functions that vanish outside the box. If the Hamiltonian is describing a two-particle system, then the Hilbert space would be the space of square-integrable functions of two variables. In general, you can think of the Hilbert space as being the space of all possible states that the system can occupy. When you have a Hamiltonian, you can usually split it up into two parts and solve them separately and then take the tensor product of the solutions to call it the solution to the total Hamiltonian. However, this is only true for certain types of Hamiltonians. For example, if the Hamiltonian describes a system with two degrees of freedom, then you can split it up into two terms corresponding to each degree of freedom and solve them separately. However, if the Hamiltonian describes a three-particle system, then you cannot do this since the interactions between the particles will mean that the Hamiltonian cannot be split up into separate terms corresponding to each particle.
 

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