Transformations of Operators and Symmetry: Difference between revisions
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{{Quantum Mechanics A}} | {{Quantum Mechanics A}} | ||
==Transformations of Operators== | |||
In | In the previous section, we discussed operators as transformations of vectors. In many cases, however, we will be interested in how operators, observables in particular, will transform under the action of another operator. Given an operator <math>\hat{A}</math> and a transformation <math>\hat{T},</math> we define the transformed operator <math>\hat{A}'</math> as follows. Given the relation, | ||
<math>\hat{A}|\psi\rangle=|\phi\rangle,</math> | |||
between two vectors <math>|\psi\rangle</math> and <math>|\phi\rangle</math>, the operator <math>\hat{A}'</math> is the operator giving the relation between <math>|\psi'\rangle=\hat{T}|\psi\rangle</math> and <math>|\phi'\rangle=\hat{U}|\phi\rangle;</math> i.e., | |||
<math>\hat{A}'|\psi'\rangle=|\phi'\rangle.</math> | |||
To find <math>\hat{A}',</math> let us first act on both sides of the original relation with <math>\hat{T}:</math> | |||
<math>\hat{T}\hat{A}|\psi\rangle=\hat{T}|\phi\rangle</math> | |||
<math> | We now introduce the identity between <math>\hat{A}</math> and <math>|\psi\rangle</math> in the form, <math>\hat{T}^{-1}\hat{T}:</math> | ||
<math>\hat{T}\hat{A}\hat{T}^{-1}\hat{T}|\psi\rangle=\hat{T}|\phi\rangle</math> | |||
<math> | | Using the above definitions of <math>|\psi'\rangle</math> and <math>|\phi'\rangle,</math> we may write this as | ||
<math>\hat{T}\hat{A}\hat{T}^{-1}|\psi'\rangle=|\phi'\rangle</math> | |||
We see then that the transformed operator <math>\hat{A}'=\hat{T}\hat{A}\hat{T}^{-1}.</math> In matrix form, this would simply correspond to a similarity transformation of <math>\hat{A}.</math> | |||
<math>\ | Of particular importance is the case in which <math>\hat{T}</math> is unitary and <math>\hat{A}</math> is an observable. This is because, in addition to preserving the normalization of the state vectors, as mentioned in the previous section, they also preserve the Hermitian nature of <math>\hat{A}:</math> | ||
<math>\hat{A}'^{\dagger}=(\hat{T}\hat{A}\hat{T}^\dagger)^\dagger=\hat{T}\hat{A}^\dagger\hat{T}^\dagger=\hat{T}\hat{A}\hat{T}^\dagger=\hat{A}'</math> | |||
== Symmetry and its Role in Quantum Mechanics == | |||
can be | Having discussed the transformation of operators, we will now apply our results to discuss symmetries of the Hamiltonian, a very important topic. As alluded to in the previous section, identifying the symmetries of the Hamiltonian will allow us to greatly simplify the problem at hand. In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities via Noether's Theorem. Moreover, in quantum mechanics, the importance of symmetries is further enhanced by the fact that measurements of conserved quantities can be exact in spite of the probabilistic nature of quantum predictions. | ||
Given a unitary transformation <math>\hat{U},</math> we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., if <math>\hat{H}'=\hat{U}\hat{H}\hat{U}^\dagger=\hat{H}.</math> We will now show that, if a transformation is a symmetry of the Hamiltonian, then it commutes with the Hamiltonian. To see this, let us take the relation, | |||
<math>\ | <math>\hat{H}|\psi\rangle=|\phi\rangle,</math> | ||
and act on both sides with <math>\hat{U}:</math> | |||
<math>\hat{U}\hat{H}|\psi\rangle=\hat{U}|\phi\rangle</math> | |||
<math>\ | Now, if <math>\hat{U}</math> is a symmetry of the Hamiltonian, then it must also be true that | ||
<math>\hat{H}\hat{U}|\psi\rangle=\hat{U}|\phi\rangle.</math> | |||
Subtracting these two equations, we see that, because <math>|\psi\rangle</math> is arbitrary, the Hamiltonian commutes with the transformation operator; i.e., <math>[\hat{H},\hat{U}]=0.</math> | |||
[[Phy5645/symmetryprob1|Problem on symmetry]] | [[Phy5645/symmetryprob1|Problem on symmetry]] |
Revision as of 16:20, 12 July 2013
Transformations of Operators
In the previous section, we discussed operators as transformations of vectors. In many cases, however, we will be interested in how operators, observables in particular, will transform under the action of another operator. Given an operator and a transformation we define the transformed operator as follows. Given the relation,
between two vectors and , the operator is the operator giving the relation between and i.e.,
To find let us first act on both sides of the original relation with
We now introduce the identity between and in the form,
Using the above definitions of and we may write this as
We see then that the transformed operator In matrix form, this would simply correspond to a similarity transformation of
Of particular importance is the case in which is unitary and is an observable. This is because, in addition to preserving the normalization of the state vectors, as mentioned in the previous section, they also preserve the Hermitian nature of
Symmetry and its Role in Quantum Mechanics
Having discussed the transformation of operators, we will now apply our results to discuss symmetries of the Hamiltonian, a very important topic. As alluded to in the previous section, identifying the symmetries of the Hamiltonian will allow us to greatly simplify the problem at hand. In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities via Noether's Theorem. Moreover, in quantum mechanics, the importance of symmetries is further enhanced by the fact that measurements of conserved quantities can be exact in spite of the probabilistic nature of quantum predictions.
Given a unitary transformation we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., if We will now show that, if a transformation is a symmetry of the Hamiltonian, then it commutes with the Hamiltonian. To see this, let us take the relation,
and act on both sides with
Now, if is a symmetry of the Hamiltonian, then it must also be true that
Subtracting these two equations, we see that, because is arbitrary, the Hamiltonian commutes with the transformation operator; i.e.,
Commutators & symmetry
We can define an operator called the parity operator, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P}} which does the following:
The parity operator commutes with the Hamiltonian if the potential is symmetric, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{V}(r)=\hat{V}(-r)} . Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator. This means that if the potential is symmetric, the solutions can be chosen to have definite parity (even and odd functions).