Partial transposition in a finite-dimensional Hilbert space: physical interpretation, measurement of observables, and entanglement

Abstract

We show that partial transposition for pure and mixed two-particle states in a discrete N-dimensional Hilbert space is equivalent to a change in sign of a “momentum-like” variable of one of the particles in the Wigner function for the state. This generalizes a result obtained for continuous-variable systems to the discrete-variable system case. We show that, in principle, quantum mechanics allows measuring the expectation value of an observable in a partially transposed state, in spite of the fact that the latter may not be a physical state. We illustrate this result with the example of an “isotropic state”, which is dependent on a parameter r, and an operator whose variance becomes negative for the partially transposed state for certain values of r; for such rs, the original states are entangled.

Appendices

A Schwinger operators for one particle

We consider an N-dimensional Hilbert space spanned by N distinct states $|q\rangle$, with $q=0,1, \ldots ,(N-1)$, which are subject to the periodic condition $|q+N\rangle =|q\rangle$. These states are designated as the “reference basis” of the space. We follow Schwinger [17, 18] and introduce the unitary operators ${\hat{X}}$ and ${\hat{Z}}$, defined by their action on the states of the reference basis by the equations:

$$\begin{aligned} {\hat{Z}}|q\rangle= & {} \omega ^q \, |q\rangle , \quad \omega =e^{2 \pi i/N}, \end{aligned}$$

(60)

$$\begin{aligned} {\hat{X}}|q\rangle= & {} |q+1\rangle . \end{aligned}$$

(61)

The operators ${\hat{X}}$ and ${\hat{Z}}$ fulfill the periodicity condition:

$$\begin{aligned} {\hat{X}}^N = {\hat{Z}}^N = \hat{{\mathbb {I}}}, \end{aligned}$$

(62)

$\hat{{\mathbb {I}}}$ being the unit operator. These definitions lead to the commutation relation:

$$\begin{aligned} {\hat{Z}}{\hat{X}}=\omega \, {\hat{X}}{\hat{Z}}. \end{aligned}$$

(63)

The two operators ${\hat{Z}}$ and ${\hat{X}}$ form a complete algebraic set, in that only a multiple of the identity commutes with both [17, 18]. As a consequence, any operator defined in our N-dimensional Hilbert space can be written as a function of ${\hat{Z}}$ and ${\hat{X}}$. We also introduce (i.e., define) the Hermitian operators ${\hat{p}}$ and ${\hat{q}}$, which play the role of “momentum” and “position”, through the equations [16, 19]:

$$\begin{aligned} {\hat{X}}= & {} \omega ^{-{\hat{p}}} = e^{-\frac{2\pi i}{N}{\hat{p}}} \; , \end{aligned}$$

(64)

$$\begin{aligned} {\hat{Z}}= & {} \omega ^{{\hat{q}}} =e^{\frac{2\pi i}{N}{\hat{q}}}. \end{aligned}$$

(65)

What we defined as the reference basis can thus be considered as the “position basis”. With (63) and definitions (64), (65), the commutator of ${\hat{q}}$ and ${\hat{p}}$ in the continuous limit [16, 19] is the standard one, $[{\hat{q}},{\hat{p}}]=i$.

B Proof of Eq. (3)

The joint probability distribution of the two momenta $p_1, p_2$ in the state ${\hat{\rho }}$ is given by

$$\begin{aligned} {{\mathcal {P}}}_{{\hat{\rho }}}\left( p_1,p_2\right)= & {} {\mathrm {Tr}} \left( {\hat{\rho }} \; {\mathbb {P}}_{p_1} \otimes {\mathbb {P}}_{p_2}\right) \end{aligned}$$

(66)

$$\begin{aligned}= & {} \sum _{n_1 n_2 n_1'n_2'} \left\langle n_1, n_2 | {\hat{\rho }} | n_1', n_2' \right\rangle \left\langle n_1', n_2'| {\mathbb {P}}_{p_1} \otimes {\mathbb {P}}_{p_2}|n_1, n_2 \right\rangle \end{aligned}$$

(67)

$$\begin{aligned}= & {} \frac{1}{N^2}\sum _{n_1 n_2 n_1'n_2'} \left\langle n_1, n_2 | {\hat{\rho }} | n_1', n_2' \right\rangle \omega ^{p_1\left( n_1'-n_1\right) } \omega ^{p_2\left( n_2'-n_2\right) }. \end{aligned}$$

(68)

The joint probability distribution of the two momenta $p_1, p_2$ for the PT operator ${\hat{\rho }}^{T_1}$ is given by

$$\begin{aligned} {{\mathcal {P}}}_{{\hat{\rho }}^{T_1}}\left( p_1,p_2\right)= & {} {\mathrm {Tr}} \left( {\hat{\rho }}^{T_1} {\mathbb {P}}_{p_1} \otimes {\mathbb {P}}_{p_2}\right) \end{aligned}$$

(69)

$$\begin{aligned}= & {} \sum _{n_1 n_2 n_1'n_2'} \left\langle n_1, n_2 | {\hat{\rho }}^{T_1} | n_1', n_2' \right\rangle \left\langle n_1', n_2'| {\mathbb {P}}_{p_1} \otimes {\mathbb {P}}_{p_2}|n_1, n_2 \right\rangle \end{aligned}$$

(70)

$$\begin{aligned}= & {} \frac{1}{N^2}\sum _{n_1 n_2 n_1'n_2'} \left\langle n_1, n_2 | {\hat{\rho }} | n_1', n_2' \right\rangle \omega ^{-p_1\left( n_1'-n_1\right) } \omega ^{p_2\left( n_2'-n_2\right) } \end{aligned}$$

(71)

$$\begin{aligned}= & {} {{\mathcal {P}}}_{{\hat{\rho }}}\left( -p_1,p_2\right) . \end{aligned}$$

(72)

This proves Eq. (3). The above proof applies for $N>2$, since, for $N=2$, $| p\rangle = |-p\rangle$.

For the case of only one particle, the above result reduces to that of Eq. (1).

C Proof of Eqs. (2) and (4)

We define the Wigner function for the density operator ${\hat{\rho }}$ as in Ref. [12], as

$$\begin{aligned} W_{{\hat{\rho }}}\left( q_1,q_2,p_1,p_2\right) = {\mathrm {Tr}}\left[ {\hat{\rho }}\left( {\hat{P}}_{q_1 p_1}\otimes {\hat{P}}_{q_2 p_2}\right) \right] \; , \end{aligned}$$

(73)

where ${\hat{P}}_{q_i p_i}$ is the “line operator” for particle i, also defined in the above reference. Explicitly, we find

$$\begin{aligned} W_{{\hat{\rho }}}\left( q_1,q_2,p_1,p_2\right) = \sum _{q_1' q_2' q_1'' q_2''} \left\langle q_1', q_2'| {\hat{\rho }} |q_1'', q_2'' \right\rangle \delta _{q_1''+q_1',2q_1} \omega ^{p_1 \left( q_1''-q_1'\right) } \delta _{q_2''+q_2',2q_2} \omega ^{p_2 \left( q_2''-q_2'\right) }. \end{aligned}$$

(74)

By definition, the Wigner function after $\hbox {PT}_{1}$ is then

$$\begin{aligned} W_{{\hat{\rho }}^{T_1}}\left( q_1,q_2,p_1,p_2\right)= & {} \sum _{q_1' q_2' q_1'' q_2''} \left\langle q_1'', q_2'| {\hat{\rho }} |q_1', q_2'' \right\rangle \delta _{q_1''+q_1',2q_1} \omega ^{p_1 \left( q_1''-q_1'\right) } \delta _{q_2''+q_2',2q_2} \omega ^{p_2 \left( q_2''-q_2'\right) } \end{aligned}$$

(75)

$$\begin{aligned} \left( q_1'\Leftrightarrow q_1''\right)= & {} \sum _{q_1' q_2' q_1'' q_2''} \left\langle q_1', q_2'| {\hat{\rho }} |q_1'', q_2'' \right\rangle \delta _{q_1'+q_1'',2q_1} \omega ^{-p_1 \left( q_1''-q_1'\right) } \delta _{q_2''+q_2',2q_2} \omega ^{p_2 \left( q_2''-q_2'\right) } \end{aligned}$$

(76)

$$\begin{aligned}= & {} W_{{\hat{\rho }}}\left( q_1,q_2,-p_1,p_2\right) . \end{aligned}$$

(77)

This proves Eq. (4).

For the case of only one particle, the above result reduces to that of Eq. (2).

D Proof of Eqs. (56), (57), and (58)

From the properties of one-particle Schwinger operators summarized in Appendix A, one can prove the following identities:

$$\begin{aligned} \frac{1}{N}{\hbox {Tr}}\left[ ({\hat{X}}^m{\hat{Z}}^l)\left( {\hat{X}}^{m'}{\hat{Z}}^{l'}\right) ^{\dagger }\right]= & {} \delta _{mm'}\delta _{ll'} \; , \end{aligned}$$

(78)

$$\begin{aligned} \frac{1}{N}{\hbox {Tr}}\left[ ({\hat{X}}^m{\hat{Z}}^l)\left( X^{m'}{\hat{Z}}^{l'}\right) ^{\dagger } \left( {\hat{X}}^m{\hat{Z}}^l \right) ^{\dagger }({\hat{X}}^{m'}{\hat{Z}}^{l'})\right]= & {} \omega ^{ml'-m'l}. \end{aligned}$$

(79)

We write the PT of the state of Eq. (47) as

$$\begin{aligned} {\hat{\rho }}_r^{T_1}= & {} \frac{r}{N} \sum _{q,q'} |q'q\rangle \langle q q'| + \frac{1-r}{N^2} \sum _{q_1,q_2} |q_1 q_2\rangle \langle q_1 q_2|, \end{aligned}$$

(80)

$$\begin{aligned}\equiv & {} r {\hat{\rho }}' +(1-r){\hat{\rho }}^{''}. \end{aligned}$$

(81)

For the first moment of ${\hat{{\varOmega }}}$, we then find

$$\begin{aligned} \hbox {Tr}({\hat{{\varOmega }}} {\hat{\rho }}')= & {} \frac{1}{N} \sum _{q,q'm,l} x_{ml} \times \left\langle q\left| {\hat{X}}_1^m {\hat{Z}}_1^l \right| q'\right\rangle {_{_1}} \times \left\langle q'\left| \left( {\hat{X}}_2^m {\hat{Z}}_2^l\right) ^{\dagger } \right| q \right\rangle _2 \nonumber \\= & {} \frac{1}{N} \sum _{q,q'm,l} x_{ml} \left\langle q\left| {\hat{X}} ^m {\hat{Z}}^l \right| q'\right\rangle \left\langle q'\left| ({\hat{X}}^m {\hat{Z}}^l)^{\dagger } \right| q\right\rangle \nonumber \\= & {} \frac{1}{N} \sum _{m,l} x_{ml}\; \hbox {Tr}\left[ ({\hat{X}}^m {\hat{Z}}^l)({\hat{X}}^m {\hat{Z}}^l)^{\dagger }\right] \nonumber \\= & {} \sum _{m,l} x_{ml} \end{aligned}$$

(82)

$$\begin{aligned} \hbox {Tr}({\hat{{\varOmega }}} {\hat{\rho }}{''})= & {} \frac{1}{N^2}\sum _{m,l} x_{ml} \hbox {Tr}\left[ \left( \hat{X_1}^m\hat{Z_1}^l\right) \left( \hat{X_2}^m\hat{Z_2}^l\right) ^{\dagger }\right] \nonumber \\= & {} \sum _{m,l} x_{ml} \delta _{m0}\delta _{l0} =x_{00}. \end{aligned}$$

(83)

We used the identity (78) to obtain Eq. (83). Equations (82) and (83) are used to prove Eq. (56) in the text.

For the second moment of ${\hat{{\varOmega }}}$, we have

$$\begin{aligned}&\hbox {Tr}({\hat{{\varOmega }}} {\hat{{\varOmega }}}^{\dagger } {\hat{\rho }}') \end{aligned}$$

(84)

$$\begin{aligned}&\quad = \sum _{m,l,m',l'} x_{ml} x_{m'l'}^* \frac{1}{N}\sum _{qq'} \left\langle q \left| ({\hat{X}}_1^m{\hat{Z}}_1^l)\left( {\hat{X}}_1^{m'}{\hat{Z}}_1^{l'}\right) ^{\dagger } \right| q'\right\rangle _1 \times \left\langle q'\left| \left( {\hat{X}}_2^m {\hat{Z}}_2^l\right) ^{\dagger }({\hat{X}}_2^{m'}{\hat{Z}}_2^{l'})\right| q\right\rangle _2 \nonumber \\&\quad = \sum _{m,l,m',l'} x_{ml} x_{m'l'}^* \frac{1}{N}\sum _{qq'} \left\langle q \left| ({\hat{X}}^m{\hat{Z}}^l)\left( {\hat{X}}^{m'}{\hat{Z}}^{l'}\right) ^{\dagger } \right| q'\right\rangle \left\langle q'\left| \left( {\hat{X}}^m {\hat{Z}}^l\right) ^{\dagger }({\hat{X}}^{m'}{\hat{Z}}^{l'})\right| q\right\rangle \nonumber \\&\quad = \sum _{m,l,m',l'} x_{ml}x_{m'l'}^* \frac{1}{N}{\hbox {Tr}}\left[ ({\hat{X}}^m{\hat{Z}}^l) \left( {\hat{X}}^{m'}{\hat{Z}}^{l'}\right) ^{\dagger } \left( {\hat{X}}^m{\hat{Z}}^l \right) ^{\dagger }({\hat{X}}^{m'}{\hat{Z}}^{l'}) \right] , \nonumber \\&\quad =\sum _{m,l,m',l'} x_{ml} x_{m'l'}^* \; \omega ^{ml'-m'l}. \end{aligned}$$

(85)

$$\begin{aligned} {\hbox {Tr}}({\hat{{\varOmega }}} {\hat{{\varOmega }}}^{\dagger } {\hat{\rho }}'')= & {} \sum _{m,l,m',l'} x_{ml} x_{m'l'}^* \frac{1}{N} {\hbox {Tr}_1}\left[ \left( {\hat{X}}_1^m{\hat{Z}}_1^l\right) \left( {\hat{X}}_1^{m'}{\hat{Z}}_1^{l'}\right) ^{\dagger }\right] \frac{1}{N} {\hbox {Tr}}_2\left[ \left( {\hat{X}}_2^{m}{\hat{Z}}_2^{l}\right) ^{\dagger }\left( {\hat{X}}_2^{m'}{\hat{Z}}_2^{l'}\right) \right] \nonumber \\= & {} \sum _{m,l,m',l'} x_{ml} x_{m'l'}^* \delta _{mm'} \delta _{ll'} = \sum _{m,l} |x_{ml}|^2 \end{aligned}$$

(86)

To obtain Eq. (85), we made use of the identity (79), and to obtain Eq. (86), we made use of the identity (78). Equations (85) and (86) are used to prove Eq. (57) in the text.

From Eqs. (57) and (56), we find Eq. (58) for the variance.