peterarose
commented
2 years ago If I understand your question correctly, you're wondering why I specify a calculation_size that is larger than the desired size of the operators. The reason is that the ladder operators are (in principle) infinite in size, and we have to truncate them to work with them numerically. The truncation always leads to errors, but we can reduce the number of errors by defining x using a slightly larger size. That way, when we calculate things like x^2, x^3, etc. they are correct for the size that is actually desired. If you set calculation_size = size, and request x^2, the [-1,-1] entry will have an error. You can compare to the same result where calculation_size = size+1.
Has default value bigger than selected on for define other operator Why .... I tested it and doesn't give any further modifications for truncated hilbert space