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Gere

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- TL;DR Summary
- An attempt to do quantum mechanics in no more than a real vector space

Is the following a correct demonstration that quantum mechanics can be done in a real vector space?

If you simply stack the entries of density matrices into a column vector, then the expression ##\textrm{Tr}(AB^\dagger)## is the same as the dot product in a complex vector space (Frobenius inner product). Therefore density matrices as objects can been seen as vectors with an inner product. Of course, not all vectors are valid density matrices, but only those with trace 1. However, as we will be interested only in inner products or convex combinations, we are not concerned about other vectors (hence "part of" a vector space).

Since the density matrix is Hermitian, and entries opposite of the diagonal are conjugates, we can change the basis with a similarity transformation to get an all real matrix. The easiest way to do that is to consider only the upper triangular and for the off diagonal terms we simply concatenate ##\sqrt{2}\Re(x_{ij})## and ##\sqrt{2}\Im(x_{ij})## to our column vector instead of ##x_{ij}## and ##x_{ji}## which were complex. Think of using the basis ##|x_i\rangle\langle x_j|+|x_j\rangle\langle x_i|## and ##i(|x_i\rangle\langle x_j|-|x_j\rangle\langle x_i|)## and normalize.

Therefore the density matrix can be a real matrix in some basis. If we take an observable eigenvector and use ##|\lambda\rangle\langle\lambda|## in our inner product, we can see that the probability for observing this eigenvector is an inner product ##P(\lambda)=\rho \cdot \lambda##. Here ##|\lambda\rangle\langle\lambda|## is like an operator where the observable value of ##|\lambda\rangle## is 1 and 0 for all other eigenvectors so that we calculate the expectation value. We use the same basis so that it will be real. Note that obserable outcomes (eigenvalues of the observable operator) are not used.

We conclude that to project out probabilities of observables, we can use a real vector space and an inner product. As an example, a single spin in this notation would simply be $$\psi_{\hat{n}}=\frac{1}{\sqrt{2}}(\hat{n}+e_4)$$. This is similar to the equation at the end of the paragraph Density matrix, but note that ##\frac{1}{\sqrt{2}}(\hat{n}+e_4)## it does not need complex numbers or Pauli matrices. There is even a simple geometric way to get this solution right away. It is straightforward to see that a measurement of ##|\uparrow\rangle## in an arbitrary direction comes out as $$\frac{1}{\sqrt{2}}(\hat{n}_1+e_4)\cdot\frac{1}{\sqrt{2}}(\hat{n}_2+e_4)=\frac{1}{2}(\cos\theta +1)=\cos^2\frac{\theta}{2}$$ as required.

The only issue for now is, that the von Neumann equation for time evolution will make the density matrix complex again. However, if at each time instant we make our matrix real again, then at least as long as we stay during that time instant, we can do the observable projection part in our real vector space. We cannot combine vectors from different times as long as he have not derived a good mapping. However, there will be a mapping because we can take our real vector, build up the complex density again, evolve the density matrix and finally get back a new real vector for a later time.

Does this show that quantum mechanics can be done in a real vector space (once you have written out the form of the time evolution)?

Update: To make it more clear. Effectively I'm doing all of QM on real vectors and operators are not needed anymore. Only the time evolution will need a fix - probably geometric algebra.

If you simply stack the entries of density matrices into a column vector, then the expression ##\textrm{Tr}(AB^\dagger)## is the same as the dot product in a complex vector space (Frobenius inner product). Therefore density matrices as objects can been seen as vectors with an inner product. Of course, not all vectors are valid density matrices, but only those with trace 1. However, as we will be interested only in inner products or convex combinations, we are not concerned about other vectors (hence "part of" a vector space).

Since the density matrix is Hermitian, and entries opposite of the diagonal are conjugates, we can change the basis with a similarity transformation to get an all real matrix. The easiest way to do that is to consider only the upper triangular and for the off diagonal terms we simply concatenate ##\sqrt{2}\Re(x_{ij})## and ##\sqrt{2}\Im(x_{ij})## to our column vector instead of ##x_{ij}## and ##x_{ji}## which were complex. Think of using the basis ##|x_i\rangle\langle x_j|+|x_j\rangle\langle x_i|## and ##i(|x_i\rangle\langle x_j|-|x_j\rangle\langle x_i|)## and normalize.

Therefore the density matrix can be a real matrix in some basis. If we take an observable eigenvector and use ##|\lambda\rangle\langle\lambda|## in our inner product, we can see that the probability for observing this eigenvector is an inner product ##P(\lambda)=\rho \cdot \lambda##. Here ##|\lambda\rangle\langle\lambda|## is like an operator where the observable value of ##|\lambda\rangle## is 1 and 0 for all other eigenvectors so that we calculate the expectation value. We use the same basis so that it will be real. Note that obserable outcomes (eigenvalues of the observable operator) are not used.

We conclude that to project out probabilities of observables, we can use a real vector space and an inner product. As an example, a single spin in this notation would simply be $$\psi_{\hat{n}}=\frac{1}{\sqrt{2}}(\hat{n}+e_4)$$. This is similar to the equation at the end of the paragraph Density matrix, but note that ##\frac{1}{\sqrt{2}}(\hat{n}+e_4)## it does not need complex numbers or Pauli matrices. There is even a simple geometric way to get this solution right away. It is straightforward to see that a measurement of ##|\uparrow\rangle## in an arbitrary direction comes out as $$\frac{1}{\sqrt{2}}(\hat{n}_1+e_4)\cdot\frac{1}{\sqrt{2}}(\hat{n}_2+e_4)=\frac{1}{2}(\cos\theta +1)=\cos^2\frac{\theta}{2}$$ as required.

The only issue for now is, that the von Neumann equation for time evolution will make the density matrix complex again. However, if at each time instant we make our matrix real again, then at least as long as we stay during that time instant, we can do the observable projection part in our real vector space. We cannot combine vectors from different times as long as he have not derived a good mapping. However, there will be a mapping because we can take our real vector, build up the complex density again, evolve the density matrix and finally get back a new real vector for a later time.

Does this show that quantum mechanics can be done in a real vector space (once you have written out the form of the time evolution)?

Update: To make it more clear. Effectively I'm doing all of QM on real vectors and operators are not needed anymore. Only the time evolution will need a fix - probably geometric algebra.

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