It is known that for time reversal operation $T$, $T|\vec r\rangle=|\vec r\rangle$ and $T|\vec p\rangle=|\vec -p\rangle$. Considering $\langle T \vec r|T\vec p \rangle$.

  1. $\langle T \vec r|T\vec p \rangle= \langle \vec r|\vec p \rangle^*$

  2. $\langle T \vec r|T\vec p \rangle= \langle \vec r|-\vec p \rangle$.

However both 1 and 2 transform $e^{-i(\mathbf{p} \cdot \mathbf{r} )}$ to $e^{+i(\mathbf{p} \cdot \mathbf{r} )}$ According to 1 the time reversal operation is due to complex conjugation but from 2 it is the operation of reversal of $\vec p$. Which interpretation of time reversal operation is correct?


1 Answer 1


(For the sake of being concise and consistent with the question, this answer is expressed using "eigenvectors" of the position and momentum operators, even though these "eigenvectors" are not normalizable and don't really belong to the Hilbert space. This caveat does not affect the essence of the question or the answer.)

The time-reversal operator $T$ is anti-linear. In general, an anti-linear operator $T$ satisfies $$ Tz|\psi\rangle=z^*T|\psi\rangle $$ for all complex numbers $z$ and all state-vectors $|\psi\rangle$. However, this by itself is not enough to specify the operator $T$. To specify the effect of an anti-linear operator on arbitrary state-vectors, we need to specify its effect on a specific basis, and then its effect on all other state-vectors follows automatically.

If we define an anti-linear operator $T$ to preserve the position eigenvectors $|\mathbf{r}\rangle$, so that $$ T|\mathbf{r}\rangle = |\mathbf{r}\rangle \tag{1} $$ for every $\mathbf{r}$, then anti-linearity immediately implies that its effect on an arbitrary state-vector is $$ T\int d^3r\ \psi(\mathbf{r})|\mathbf{r}\rangle =\int d^3r\ \psi^*(\mathbf{r})|\mathbf{r}\rangle. \tag{2} $$ In particular, its effect on the momentum eigenfunctions $$ |\mathbf{p}\rangle = \int d^3r\ \exp(i\mathbf{p}\cdot\mathbf{r})|\mathbf{r}\rangle \tag{3} $$ is $$ T\int d^3r\ \exp(i\mathbf{p}\cdot\mathbf{r})|\mathbf{r}\rangle =\int d^3r\ \exp(-i\mathbf{p}\cdot\mathbf{r})|\mathbf{r}\rangle, \tag{4} $$ which may also be written $$ T|\mathbf{p}\rangle = |-\mathbf{p}\rangle. \tag{5} $$ This is consistent with both equations 1 and 2 in the question.

The key here is that $T$ acts by complex conjugation only in the position basis, as shown in equation (2). It does not act as complex conjugation in other bases. In particular, it does not act as complex conjugation in the momentum basis. Equation (5) implies that the effect of $T$ on an arbitrary state-vector expressed in the momentum basis is \begin{align} T \int \frac{d^3p}{(2\pi)^3}\ \phi(\mathbf{p})|\mathbf{p}\rangle &= \int \frac{d^3p}{(2\pi)^3}\ \phi^*(\mathbf{p})T|\mathbf{p}\rangle \\ &= \int \frac{d^3p}{(2\pi)^3}\ \phi^*(\mathbf{p})|-\mathbf{p}\rangle \\ &= \int \frac{d^3p}{(2\pi)^3}\ \phi^*(-\mathbf{p})|\mathbf{p}\rangle. \tag{6} \end{align} Since $\phi^*(-\mathbf{p})\neq\phi^*(\mathbf{p})$ in general, this shows that $T$ does not act as complex conjugation in general; that would lead to contradictions. Another way to deduce (6) is to substitute $$ |\mathbf{r}\rangle = \int \frac{d^3p}{(2\pi)^3}\ \exp(-i\mathbf{p}\cdot\mathbf{r})|\mathbf{p}\rangle \tag{7} $$ into equation (2). Since $T$ was defined to leave $|\mathbf{r}\rangle$ invariant, it cannot merely take the complex conjugate of the coefficient $\exp(-i\mathbf{p}\cdot\mathbf{r})$ in equation (7). The net effect of $T$ is described by equation (6).

So the answer to the question is that time reversal acts by complex conjugation in the position basis only. Equations 1 and 2 in the question are both correct, and in fact they are saying the same thing because the right-hand sides of those equation are equal to each other. Strictly speaking, equation 2 in the question doesn't say that $T$ reverses the parameter $\mathbf{p}$. It says that $T$ replaces the state-vector $|\mathbf{p}\rangle$ with the state-vector $|-\mathbf{p}\rangle$, which is the same as saying that $T$ acts as complex conjugation in the position basis.


Here's a simpler illustration of the fact that an antilinear operator (like the time-reversal operator) acts by complex conjugation only in one basis and not in other bases.

Every antilinear operator satisfies $$ Tz|\psi\rangle = z^*T|\psi\rangle \tag{A-1} $$ for every state-vector $|\psi\rangle$ and every complex number $z$. This is what "antilinear" means. Let $|0\rangle$ be any state-vector, and let $T$ be an antilinear operator $T$ that satisfies $$ T|0\rangle = |0\rangle. \tag{A-2} $$ Now let $\theta$ be a real variable and define $$ |\theta\rangle \equiv \exp(i\theta)|0\rangle. \tag{A-3} $$ Equations (A-1) and (A-2) imply $$ T|\theta\rangle = \exp(-i\theta)T|0\rangle = \exp(-i\theta)|0\rangle=|-\theta\rangle. \tag{A-4} $$ Therefore, $$ Tz|\theta\rangle =z^*T|\theta\rangle =z^*|-\theta\rangle \tag{A-5} $$ For most values of $\theta$, this implies $$ Tz|\theta\rangle \neq z^*|\theta\rangle. \tag{A-6} $$ This demonstrates once again that $T$ does not always act by complex conjugating the coefficients. It can only act that way in one basis.

A "wavefunction" is a set of coefficients. For example, in the state-vector $$ \int d^3r\ \psi(\mathbf{r})|\mathbf{r}\rangle = \int \frac{d^3p}{(2\pi)^3}\ \phi(\mathbf{p})|\mathbf{p}\rangle \tag{A-7} $$ with $$ \phi(\mathbf{p})\equiv \int d^3r\ \exp(-i\mathbf{p}\cdot\mathbf{r})\psi(\mathbf{r}), \tag{A-8} $$ the coefficients $\psi(\mathbf{r})$ constitute the wavefunction in the position basis, and the coefficients $\phi(\mathbf{p})$ constitute the wavefunction in the momentum basis. Both represent the same state-vector. $T$ acts by complex conjugation of $\psi(\mathbf{r})$, but $T$ does not act by complex conjugation of $\phi(\mathbf{p})$. This was demonstrated by equations (2) and (6) in the main text, and equations (A-1) to (A-6) illustrate the same phenomenon more simply.

  • $\begingroup$ I am having some difficulty in trying to understand your answer. Will be thankful for any help. Actually time reversal can be written as a product of unitary operator and complex congugation operator.In the coordinate space the unitary operator is just an identity operator so one just takes the complex congugation. Time reversal is an operation with specefic physical significance as such it is difficult to understand how 6 and 10 can be both valid at the same time. Could you elaborate on the delta functions. They seem to be scalar products. how 5 from 4? Also your eqs.1 and 2 are missing. $\endgroup$
    – SAKhan
    Dec 2, 2018 at 11:09
  • $\begingroup$ @SAKhan Actually, after seeing your comment and re-reading my answer, even I had difficulty understanding it! So I rewrote it using state-vector notation instead of wavefunction notation. Hopefully this is more clear. Equations 1 and 2 were missing because I was continuing the numbering that was started in the question; the new version of the answer starts with number 1. In the original answer, I was writing $\delta^3(\mathbf{r}-\mathbf{c})$ to mean the wavefunction for the position eigenvector $|\mathbf{c}\rangle$ expressed in the position basis $|\mathbf{r}\rangle$ ... Ugh! Pretty confusing. $\endgroup$ Dec 3, 2018 at 0:55
  • $\begingroup$ Thanks Dan Yand. The functions in 3 and 7 are same except for the sign in the exponent. The effect of time reversal on the same functional type lead to different result. The former exponent changes sign but the latter exponent sign remains unchanged. Time reversal leading to different results. $\endgroup$
    – SAKhan
    Dec 4, 2018 at 10:26
  • $\begingroup$ @SAKhan Equations (3) and (7) are simply each other's inverses: either one implies the other, independently of how $T$ is defined. Applying $T$ to (3) and using (1) with $Tz|\psi\rangle=z^*T|\psi\rangle$ gives (5). Equivalently, applying $T$ to (7) and using (5) with $Tz|\psi\rangle=z^*T|\psi\rangle$ gives (1). So the equations are all consistent with each other, and I don't see how they can lead to different results. Can you show an example of how they lead to different results? $\endgroup$ Dec 4, 2018 at 13:34
  • $\begingroup$ @SAKhan I added an appendix to try to help clarify things. $\endgroup$ Dec 4, 2018 at 14:06

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