_{Disclaimer: I lay no claim to the originality of the basic ideas expounded here. Its distinctness lies more in its scope and in the detail of its exposition.} ![QuantumMechanics.jpg](https://cdn.steemitimages.com/DQmTspC1Cik9wck8tBBZuqyYf1w5mznES2qp7ucwiQ3Rmha/QuantumMechanics.jpg) "Let's do some real science" ## Expectation Values of Observables In quantum mechanical sense, what does the expectation value of an observable *A* mean? The **expectation value** of an observable *A* in a particle state described by a wave function ![](https://cdn.steemitimages.com/DQmZKyzhNVkKF2houKkx77xHGy7mNEs8pkyTrskvatx9AvP/image.png) is defined as the average result of **N**, as it approaches a large number, of measurements of A performed on a particle in the same state ![](https://cdn.steemitimages.com/DQmZKyzhNVkKF2houKkx77xHGy7mNEs8pkyTrskvatx9AvP/image.png). The prescription for calculating the expectation value of a normalized wave function ![](https://cdn.steemitimages.com/DQmZKyzhNVkKF2houKkx77xHGy7mNEs8pkyTrskvatx9AvP/image.png) is known as ![](https://cdn.steemitimages.com/DQmciwsLMLDPo84qA26smGvvrLEtsWarzgLDdNR9DJk4NUX/image.png) To show that this prescription is the same to the definition of the expectation value as an average, consider the following wavefunction, ![](https://cdn.steemitimages.com/DQmZYYGzpB4LNBwKiKAon2ge6C7egCmvS6SRbTHRpw81x3B/image.png) which we have covered on the case on which ![](https://cdn.steemitimages.com/DQmbggPgAiY5DZdPxMARnt7bJNRqPVnwdB94a5yAuDfYAvS/image.png) is not an eigenfunction of the operator. We substitute this equation to the prescription of the expectation value, ![](https://cdn.steemitimages.com/DQmTumwhsLejUKpPb3YsXYadu16uLujABbPZYan1uv8cjio/image.png) This relation asserts that the expectation value of *A* in the state ![](https://cdn.steemitimages.com/DQmTVcCLSmCMkhCHRLXqpewgkYhqzdpfRbzPT343M1UMLFW/image.png) is the sum of all possible outcome ![](https://cdn.steemitimages.com/DQmXqgHybaVaQQd7zycwTTxyes2aNZGTWkypoSb8j4yujQJ/image.png), each weighted by the probability ![](https://cdn.steemitimages.com/DQmexzD6m5QrAnYSFhdJUMCdh7fccnzvR795PNxdZeZVAFR/image.png) that in this state a measurement of *A* will yield the result ![](https://cdn.steemitimages.com/DQmXqgHybaVaQQd7zycwTTxyes2aNZGTWkypoSb8j4yujQJ/image.png). ## Commuting observables and simultaneous observability If the operators corresponding to two observables *A* and *B* have a common, complete set of eigenfunctions ![](https://cdn.steemitimages.com/DQmRx1C5HHVXP6FdWovW9oBouxezZ62GQUXk6px5QXKrPCJ/image.png), where we have the *i* to label the corresponding eigenvalues ![](https://cdn.steemitimages.com/DQmUB7f4HUhgHofW7NKTeKPKXkmdbeBesjsN8Jpuv2gJZV8/image.png) of ![](https://cdn.steemitimages.com/DQmVD2LMkzbGGRonszY6H2fjzyaGPdVpBvm4p81tSnTK6wN/image.png) and *j* labels the corresponding eigenvalues ![](https://cdn.steemitimages.com/DQmVi3SpsaWPWBK7ijPQwba14kNGNSSKR6mwjXyUnscUpFM/image.png) of ![](https://cdn.steemitimages.com/DQmZyZi9d2mzoMtBkJdZNJuL91vAaFcsy4NAseEAFhKk1dm/image.png), we say that *A* and *B* can be **simultaneously well defined**. Another way of saying this is that they have a simultaneously observable. **How do we know that two operators are simultaneously observable?** The commutator ![](https://cdn.steemitimages.com/DQmXuCvd32rnBu9sn6NyCw3oTY8v4wnWkxaotCDSfJkUaph/image.png) of any pair of simultaneously well-defined observables A and B turns out to be zero. This is a sufficient condition. In some sense, the commutator action on an arbitrary normalizable function ![](https://cdn.steemitimages.com/DQmXbCGaD6GsdAmYdepnm33jZnRYKTabapah5ZayMV32m9y/image.png) gives zero. Lets show this. We will use that fact that any such function ![](https://cdn.steemitimages.com/DQmXbCGaD6GsdAmYdepnm33jZnRYKTabapah5ZayMV32m9y/image.png) can be written as a linear combination of the common eigenfunctions ![](https://cdn.steemitimages.com/DQmcatPTA3uX4JJd7XKmNSm2UJ32bZ4pEtV5VoLY7TVc3Hj/image.png) of ![](https://cdn.steemitimages.com/DQmVdh143wdRUAtL6nyPSooq48BJYuRqLAAxgiGRv98EbgE/image.png), so that ![](https://cdn.steemitimages.com/DQmaAaHdacbNPAojzkuhFhAwcKyePhY1BLrcPrWBnuySaaD/image.png) The converse is also true, such that if ![](https://cdn.steemitimages.com/DQmNNBd5ehN1iDaXZ55bXjt6UME6w92rYyPd1kEUzYQ39L4/image.png) and ![](https://cdn.steemitimages.com/DQmdgWkaWt7gevue9VRqik8VinLjPJ9axjsTKRmXW2hctj9/image.png) commute, that is, ![](https://cdn.steemitimages.com/DQma8LPijBcds2y6d8xvqgobhkduY3g5zoD8ZcySQ8LWaaE/image.png) they are simultaneous eigenfunction. ## Non-commuting observables and the uncertainty principle If the commutator ![](https://cdn.steemitimages.com/DQmdMSAdGWbvNejgHGHPezciye6NhvQ1WLydQ2nWyRCSKWG/image.png) then the observables *A* and *B* cannot be simultaneously well defined. The uncertainty in the observable *A* in the state ![](https://cdn.steemitimages.com/DQmSyrzTcX5b2wetiSCG7pVbiyGCZkzrbNjX7bgAXBnS51h/image.png) is define by ![](https://cdn.steemitimages.com/DQmZwvap55Xq9UNhQeMcym94bsC1GY9CW6v4ey7UvqJxpcQ/image.png) We can show that the product of the uncertainties in two observables A and B is given by the inequality (**general uncertainty relation**) ![](https://cdn.steemitimages.com/DQmb6FjoiBAdveh4ACRceqbcjM8TYWgX1VbvAvy78cAnmh4/image.png) --- **Proof:** Take note of this relation that we derive from previous section, ![](https://cdn.steemitimages.com/DQmUMrXvtobRBCKDbddUYGhfbiJuqVhA9g7Jiii4mW2rPs5/image.png) With this in mind, we set ![](https://cdn.steemitimages.com/DQmXirPyopoNHJqkBjEuYGAktCVGfc2wxCPD9Npa8B15DL3/image.png). So we have, ![](https://cdn.steemitimages.com/DQmZjuGYEAASqvJHG4DfrV25YfZSvSqMd7g1bNkuVhuneyL/image.png) Using a similar line of reasoning we can show that, ![](https://cdn.steemitimages.com/DQmUZDijfoiHTDWS4QY4873rUFWQaV4JHECDdVr7MCzf7oG/image.png) Then the inequality becomes: ![](https://cdn.steemitimages.com/DQmTBjRkz7W9K7aKBuoDJUdcW6GU1tWsJ33LynBJAgVsFGy/image.png) Now what is left is to show what ![](https://cdn.steemitimages.com/DQmYRtd91P1SeHw9fdXvX3a2tZHywUmMR8wsnEruCrAMx19/image.png) And so we have the following relation, ![](https://cdn.steemitimages.com/DQmbUkg5nzQXnCyMtc27N1edyZkcosS2d8vXby2Nv6V2LyM/image.png) **Example:** Derive the uncertainty relation of the following operator: ![](https://cdn.steemitimages.com/DQmaz8Lh1g6C5gKqzQ25XnxF2XFZkXGRW1TnH4Dgnbd37sG/image.png), then we have, ![](https://cdn.steemitimages.com/DQmcdfeP85mJu1AKc6LeZQaAxUtANUtSqtenw1kPSeE8TfK/image.png) From this we can see that, ![](https://cdn.steemitimages.com/DQmV7SMymWxUiNbEpjVaKKEs3UDfarHM1RcZuRrLMNFUuLL/image.png) and so the uncertainty relation yields, ![](https://cdn.steemitimages.com/DQmbNE3GrdS2snnqMUHp318wP9MYS7efDjTCyK4fa7w8xBb/image.png) This implies that there is no 1D state ![](https://cdn.steemitimages.com/DQmeShR3FiZ5oGuAPFje69ck4RFWsxwnSsJqrrTSsTb6qVU/image.png) that exists for which the product of the uncertainties of the position and momentum observable is smaller than ![](https://cdn.steemitimages.com/DQmerTJq6EV5Mdf1M9KTPnWkpvxhXi7Y487N7a5izLdUzxP/image.png). It is found that in a one dimensional harmonic oscillator in its ground state, ![](https://cdn.steemitimages.com/DQmUSfxFGry79PuXKknV1SLCQPUKRDdUFnihSLJCYRDXDZg/image.png), the uncertainty becomes an equality, ![](https://cdn.steemitimages.com/DQmToYBQFK64grbAcffLqHpbrFt89WwFcRr6JKSCkeoXtYY/image.png) --- ## Time dependence of Expectation Values Use the following expression for the expectation value of an observable *A* , ![](https://cdn.steemitimages.com/DQmfJ1ZUqn6NWzH5tYvH62JgRDdf6nFHa7h7N9sN648Va1w/image.png) And also the time-dependent Schrodinger equation in the form, ![](https://cdn.steemitimages.com/DQmR23somw3qNVBbeKkkdp7wKPQfiu8w4Uzio4PTZCxV11u/image.png) From these equations, we can obtain an expression for the rate of change of the expectation value with time: ![](https://cdn.steemitimages.com/DQmZqJB7yyPyoENkqom3zLG3QYZzA6rSeh4n7dDtHFzGaLz/image.png) From this we can derive the following expression, ![](https://cdn.steemitimages.com/DQmYHw7BYmEVUQC7SEpSYBagg9pkfpf3WghdQh7mXvyUPC9/image.png) which is known as the **Ehrenfest's theorem**.

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