Heisenberg's matrix mechanics and the Uncertainty Principle
Continuing the exploration of various quantum mechanics formulations today I want to talk about Heisenberg's matrix mechanics. A major stumbling block for early 20th century physics was the explanation of discrete levels of energy exhibited for example by the atomic spectral lines. Classical mechanics was completely unable to provide any explanation and this crisis led to the exploration of new ideas.
One profound way of thinking about this was to strip the theory of all unobservable baggage (like the trajectory of an electron in an atom) and only talk about what can be actually tested by experiment. The discrete levels of energy would come out of the non-commutativity as follows:
1. Find n matrices \(Q_k\) and \(P_k\) such that \([P_i, P_j] = [Q_i, Q_j] = 0\) and \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \)
2. Diagonalize the Hamiltonian \(H\) by usage of the following transformation: \(S^{-1}Q_i S\) and \(S^{-1}P_j S\)
The diagonal elements are the energy levels observed in experiments and the theory seems at first to have the same explanation power as the old semiclassical formulation of quantum mechanics due to Bohr and Sommerfeld, but lacking a mechanism to predict intensities. However this is not the case, because lurking about there is the standard Hilbert space formulation and von Neumann brought it to the surface in his classical book: The Mathematical Foundations of Quantum Mechanics. (Historical note: Heisenberg used intensity information and working backwards he derived a non-commutative rule of multiplication which turned out to be the usual matrix multiplication)
The diagonalization step is nothing but the usual eigenvalue-eigenvector problem: \(H x = \lambda x\) where \(x\) is a column of \(S\) and the space of \(x\) is the usual Hilbert space. As expected \(P, Q, H\) are operators acting on this Hilbert space, and the intensities predictions arise out of projecting in the Hilbert space.
Then as now, to defend a new paradigm one needs to produce a physical justification, and Mr. Heisenberg tried to explain the commutation relationship \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \) in terms of the inability to measure simultaneously the position and momenta of a particle in what is now known as the uncertainty principle. Today this is so commonplace that it is easy to overlook some subtle points about it and I'll present two such points.
Suppose that you have a beam of electrons moving in a straight line and with various speeds. By placing this beam in a magnetic field you can separate the electrons according to their speed. Then if you place a photographic plate in front of them you can also measure the electron's position. The momentum has an uncertainty, the position has its uncertainty as well, but if you compute \(\Delta p \Delta x\) you seem to be able to beat the uncertainty principle!!! (try with the usual values of electron speed and detection resolution in an old cathode ray tube TV). However, this does not disprove the uncertainty principle and moreover Mr. Heisenberg was aware of this.
I can tell you the solution to this problem, but it is much more interesting for the reader to try to wrestle with this puzzle and explain the apparent contradiction (I'll provide Heisenberg's take on it in the next post).
The second subtle point (arising out of a advanced functional analysis) is that non-commutativity is not enough to establish the uncertainty principle. In addition one needs that:
1. the operators should be unbounded
2. the operators should either have no point spectrum or if they have a point spectrum that spectrum should lie outside the domain of the other operator.
Position and momenta respect those conditions, but spin does not and as such there is no uncertainty principle for spin! But wait a minute, this sounds preposterous. The commutator: \([S_x, S_y] = i\hbar S_z\) allow us to compute \(\Delta S_x \Delta S_y\) and this is not zero! So how can we claim that there is no uncertainty principle for spin? There is a subtle difference between the spin and the position-momenta case. Can you spot it? Again' I'll present the solution next time because it is much more interesting for the reader to attempt to solve this connundrum as well.
To summarize, in the matrix mechanics formulation, both the spectra and the intensities arise out of the diagonalization step after the (infinite dimensional) matrices were selected to respect the commutation relationship. What the theory does not provide is a concrete transition mechanism (corresponding to the collapse postulate) and therefore quantum mechanics is intrinsically probabilistic. This left open the possibility to "complete" quantum mechanics and various interpretations of quantum mechanics are nothing but solution to the so-called "measurement problem". More on this in subsequent posts.
1. Find n matrices \(Q_k\) and \(P_k\) such that \([P_i, P_j] = [Q_i, Q_j] = 0\) and \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \)
2. Diagonalize the Hamiltonian \(H\) by usage of the following transformation: \(S^{-1}Q_i S\) and \(S^{-1}P_j S\)
The diagonal elements are the energy levels observed in experiments and the theory seems at first to have the same explanation power as the old semiclassical formulation of quantum mechanics due to Bohr and Sommerfeld, but lacking a mechanism to predict intensities. However this is not the case, because lurking about there is the standard Hilbert space formulation and von Neumann brought it to the surface in his classical book: The Mathematical Foundations of Quantum Mechanics. (Historical note: Heisenberg used intensity information and working backwards he derived a non-commutative rule of multiplication which turned out to be the usual matrix multiplication)
The diagonalization step is nothing but the usual eigenvalue-eigenvector problem: \(H x = \lambda x\) where \(x\) is a column of \(S\) and the space of \(x\) is the usual Hilbert space. As expected \(P, Q, H\) are operators acting on this Hilbert space, and the intensities predictions arise out of projecting in the Hilbert space.
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| Werner Heisenberg |
Then as now, to defend a new paradigm one needs to produce a physical justification, and Mr. Heisenberg tried to explain the commutation relationship \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \) in terms of the inability to measure simultaneously the position and momenta of a particle in what is now known as the uncertainty principle. Today this is so commonplace that it is easy to overlook some subtle points about it and I'll present two such points.
Suppose that you have a beam of electrons moving in a straight line and with various speeds. By placing this beam in a magnetic field you can separate the electrons according to their speed. Then if you place a photographic plate in front of them you can also measure the electron's position. The momentum has an uncertainty, the position has its uncertainty as well, but if you compute \(\Delta p \Delta x\) you seem to be able to beat the uncertainty principle!!! (try with the usual values of electron speed and detection resolution in an old cathode ray tube TV). However, this does not disprove the uncertainty principle and moreover Mr. Heisenberg was aware of this.
I can tell you the solution to this problem, but it is much more interesting for the reader to try to wrestle with this puzzle and explain the apparent contradiction (I'll provide Heisenberg's take on it in the next post).
The second subtle point (arising out of a advanced functional analysis) is that non-commutativity is not enough to establish the uncertainty principle. In addition one needs that:
1. the operators should be unbounded
2. the operators should either have no point spectrum or if they have a point spectrum that spectrum should lie outside the domain of the other operator.
Position and momenta respect those conditions, but spin does not and as such there is no uncertainty principle for spin! But wait a minute, this sounds preposterous. The commutator: \([S_x, S_y] = i\hbar S_z\) allow us to compute \(\Delta S_x \Delta S_y\) and this is not zero! So how can we claim that there is no uncertainty principle for spin? There is a subtle difference between the spin and the position-momenta case. Can you spot it? Again' I'll present the solution next time because it is much more interesting for the reader to attempt to solve this connundrum as well.
To summarize, in the matrix mechanics formulation, both the spectra and the intensities arise out of the diagonalization step after the (infinite dimensional) matrices were selected to respect the commutation relationship. What the theory does not provide is a concrete transition mechanism (corresponding to the collapse postulate) and therefore quantum mechanics is intrinsically probabilistic. This left open the possibility to "complete" quantum mechanics and various interpretations of quantum mechanics are nothing but solution to the so-called "measurement problem". More on this in subsequent posts.
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