Negative Kinetic Energy and Entanglement in Quantum Bound States
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2025
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| _version_ | 1866902211582230528 |
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| author | Ruggeri, Francesco R. |
| author_facet | Ruggeri, Francesco R. |
| contents | <p>In (1), an explanation of the quantum feature of a bound state which allows for a particle to exist in a classically forbidden region and have negative kinetic energy is given. Specifically, (1) suggests that although classical mechanics represents kinetic and potential energy by numbers, quantum mechanics treats them as operators and that this is at the heart of the above curious situation.</p> <p> In this note, we consider this quantum feature from a different point of view. We note that even if one uses an operator for kinetic energy in quantum mechanics, in the free particle case, which uses a wavefunction of exp(ipx), kinetic energy is always positive and exactly equal to its classical counterpart. Thus, we suggest that measurements occur using exp(ipx). The problem is that there is an intrinsic length hbar/p linked with exp(ipx) and that it exists for all x, even though p is a single value (i.e. there is no acceleration). This makes it usual to use exp(ipx) in a description of a bound state, but given that it is the function linked with measurement, we argue that it must be present, in fact, a variety of exp(ipx)s must be present. One must then somehow associate an average kinetic energy with each x, because an exact kinetic energy from one of the exp(ipx)s does not exist at a sharp x point. </p> <p> To do this, we suggest that quantum mechanics uses something which does not have a classical analogue, namely the entangled state, W(x) = Sum over p a(p)exp(ipx), of various free particle states. To measure the kinetic energy of a particle as an exact number, as in classical mechanics, one would have to break the entangled state, i.e. knock out a particle such that it is in an exp(ipx) state. In such a case, this particle could be anywhere because exp(ipx) holds for all x. One may see that the bound state is actually an entangled one because the kinetic energy of the knocked out particle may actually be higher than the average energy of the bound state. </p> <p> The point we make, however, is that for exp(ipx), the kinetic energy is always positive even though there is an operator -1/2m d/dx d/dx. It is only the kinetic energy of the entangled state which may be negative outside the classical turning points. An entangled state, however, is not a classical state with a single kinetic energy which is positive. As a result, we suggest that the quantum bound state is described in terms of an entangled state which does not map to a classical state. The quantum free particle, on the other hand, maps to a classical state in terms of its momentum and kinetic energy, but cannot describe a bound state by itself in the quantum problem. (Furthermore, we argue that the classical turning points as sharp points is really an average feature as the potential V(x) may be thought of as Sum over k Vk exp(ikx).)</p> <p> </p> |
| format | Recurso digital |
| id | zenodo_https___doi_org_10_5281_zenodo_14629922 |
| institution | Zenodo |
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| publishDate | 2025 |
| publisher | Zenodo |
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| spellingShingle | Negative Kinetic Energy and Entanglement in Quantum Bound States Ruggeri, Francesco R. <p>In (1), an explanation of the quantum feature of a bound state which allows for a particle to exist in a classically forbidden region and have negative kinetic energy is given. Specifically, (1) suggests that although classical mechanics represents kinetic and potential energy by numbers, quantum mechanics treats them as operators and that this is at the heart of the above curious situation.</p> <p> In this note, we consider this quantum feature from a different point of view. We note that even if one uses an operator for kinetic energy in quantum mechanics, in the free particle case, which uses a wavefunction of exp(ipx), kinetic energy is always positive and exactly equal to its classical counterpart. Thus, we suggest that measurements occur using exp(ipx). The problem is that there is an intrinsic length hbar/p linked with exp(ipx) and that it exists for all x, even though p is a single value (i.e. there is no acceleration). This makes it usual to use exp(ipx) in a description of a bound state, but given that it is the function linked with measurement, we argue that it must be present, in fact, a variety of exp(ipx)s must be present. One must then somehow associate an average kinetic energy with each x, because an exact kinetic energy from one of the exp(ipx)s does not exist at a sharp x point. </p> <p> To do this, we suggest that quantum mechanics uses something which does not have a classical analogue, namely the entangled state, W(x) = Sum over p a(p)exp(ipx), of various free particle states. To measure the kinetic energy of a particle as an exact number, as in classical mechanics, one would have to break the entangled state, i.e. knock out a particle such that it is in an exp(ipx) state. In such a case, this particle could be anywhere because exp(ipx) holds for all x. One may see that the bound state is actually an entangled one because the kinetic energy of the knocked out particle may actually be higher than the average energy of the bound state. </p> <p> The point we make, however, is that for exp(ipx), the kinetic energy is always positive even though there is an operator -1/2m d/dx d/dx. It is only the kinetic energy of the entangled state which may be negative outside the classical turning points. An entangled state, however, is not a classical state with a single kinetic energy which is positive. As a result, we suggest that the quantum bound state is described in terms of an entangled state which does not map to a classical state. The quantum free particle, on the other hand, maps to a classical state in terms of its momentum and kinetic energy, but cannot describe a bound state by itself in the quantum problem. (Furthermore, we argue that the classical turning points as sharp points is really an average feature as the potential V(x) may be thought of as Sum over k Vk exp(ikx).)</p> <p> </p> |
| title | Negative Kinetic Energy and Entanglement in Quantum Bound States |
| url | https://doi.org/10.5281/zenodo.14629922 |